The rate for the reaction, $RCl + NaOH\left( {aq} \right) \to ROH + NaCl$ is given by $rate{\rm{ }} = K_1\left[ {RCl} \right].$ The rate of the reaction is

Unaffected by increasing the temperature of the reaction

Decreased on increasing the temperature of the reaction

Doubled on doubling the concentration of $NaOH$

At room temperature, the reaction between ${\rm{NO}}$ and ${{\rm{O}}_2}$ to give ${\rm{N}}{{\rm{O}}_2}$ is fast, while that between ${\rm{CO}}$ and ${{\rm{O}}_2}$ is slow. It is due to:

The activation energy for the reaction, $2{\rm{NO}} + {{\rm{O}}_2} \to 2{\rm{N}}{{\rm{O}}_2}$ is less than $2{\rm{CO}} + {{\rm{O}}_2} \to 2{\rm{C}}{{\rm{O}}_2}$

${\rm{CO}}$ is smaller in size than that of ${\rm{NO}}$

The data for the reaction, ${\rm{ }}A + B \to C$ $\begin{array}{l} Ex\;\;\;\;\;\;\;\;\;\;\;{\left[ A \right]_0}\;\;\;\;\;\;\;\;\;{\left[ B \right]_0}\;\;\;\;\;\;\;\;\;Initial\;{\rm{ }}rate\\ \begin{array}{*{20}{l}} {1\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.012\;\;\;\;\;\;\;\;0.035\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.10}\\ {2\;\;\;\;\;\;\;\;\;\;\;\;\;0.024\;\;\;\;\;\;\;\;\;0.070\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.80}\\ {3\;\;\;\;\;\;\;\;\;\;\;\;\;0.024\;\;\;\;\;\;\;\;\;\;0.035\;\;\;\;\;\;\;\;\;\;\;\;\;0.10}\\ {4\;\;\;\;\;\;\;\;\;\;\;\;\;0.012\;\;\;\;\;\;\;\;\;0.070\;\;\;\;\;\;\;\;\;\;\;\;\;0.80} \end{array} \end{array}$ The rate law corresponds to the above data is

$rate = k\left[ A \right]{\left[ B \right]^3}$

$rate = k{\left[ A \right]^2}{\left[ B \right]^2}$

CHEMICAL KINETICS MCQs for NEET UG CHEMISTRY — Medium

1 Medium

1 of 71Question permalink

Application +4 | -1

The rate constant

\left( {k'} \right)

of one of the reaction is found to be double that of the rate constant

\left( {k''} \right)

of another reaction. Then the relationship between the corresponding activation energies of the two reaction

\left( {{{E'}_a}\;\mathrm{and}\;{{E''}_a}} \right)

can be represented as

View answer and explanation

Correct option: D

{{E'}_a} < {{E''}_a}

Option D (

{{E'}_a} < {{E''}_a}

) is correct because a higher rate constant implies a lower activation energy, according to the Arrhenius equation.

Solution approach

Correct Answer + Core Reason: ${{E'}_a} < {{E''}_a}$ . A higher rate constant implies a lower activation energy, according to the Arrhenius equation.

Step-by-Step Solution:

Step 1: Write the Arrhenius equation: $k = A e^{-E_a/RT}$

Step 2: We are given that $k' = 2k''$ .

Step 3: Since $k'$ is greater than $k''$ , the term $e^{-E_a/RT}$ must be larger for the reaction with rate constant $k'$ .

Step 4: For $e^{-E_a/RT}$ to be larger, the exponent $-E_a/RT$ must be larger (less negative).

Step 5: Therefore, $E'_a$ must be smaller than $E''_a$ .

Step 6: Final Answer: ${{E'}_a} < {{E''}_a}$

Why other options are wrong:
- Option A: This is incorrect because a higher rate constant corresponds to a lower activation energy.
- Option B: This is incorrect because there is no direct proportional relationship between the activation energies.
- Option C: This is incorrect because the rate constants are different, implying different activation energies.

Key Concepts:
- Arrhenius equation
- Activation energy

Memory Aids:
- Remember the inverse relationship between the rate constant and the activation energy.

Quick Tips:
- A higher rate constant implies a faster reaction, which requires less activation energy.

2 Medium

2 of 71Question permalink

Application +4 | -1

Temperature coefficient of a reaction is 2. When temperature is increased from

30^\circ C\;to\;100^\circ C

, rate of the reaction increases by

View answer and explanation

Correct option: A

128 times

Option A (128 times) is correct because the temperature increases by 70°C, which is 7 increments of 10°C, so the rate increases by

2^7 = 128

times.

Solution approach

Correct Answer + Core Reason: 128 times. The rate increases by a factor of 2 for every 10°C increase in temperature.

Step-by-Step Solution:

Step 1: Calculate the temperature difference: $\Delta T = 100°C - 30°C = 70°C$

Step 2: Determine the number of 10°C intervals: $n = \frac{70}{10} = 7$

Step 3: Since the temperature coefficient is 2, the rate increases by a factor of $2^n$ .

Step 4: Calculate the factor: $2^7 = 128$

Step 5: Final Answer: 128 times

Why other options are wrong:
- Option B: This is incorrect because it is the temperature increase itself, not the factor by which the rate increases.
- Option C: This is incorrect because it's not a power of 2.
- Option D: This is incorrect because it's not a power of 2.

Key Concepts:
- Temperature coefficient
- Effect of temperature on reaction rate

Memory Aids:
- Remember that the rate increases exponentially with temperature when the temperature coefficient is greater than 1.

Quick Tips:
- If the temperature coefficient is 'x', and the temperature increases by $n \times 10°C$ , the rate increases by $x^n$ .

3 Medium

3 of 71Question permalink

Understanding +4 | -1

Which is correct relation in between

\frac{{dc}}{{dt}},\frac{{dn}}{{dt}}\;and\;\frac{{dP}}{{dt}},

where

c,n\;and\;p

, represent concentration, mole and pressure terms for gaseous phase reactant

A\left( \mathrm{g} \right) \to products

?

View answer and explanation

Correct option: A

- \frac{{dc}}{{dt}} = - \frac{1}{V}\frac{{dn}}{{dt}} = - \frac{1}{{RT}}\frac{{dP}}{{dt}}

Option A (

- \frac{{dc}}{{dt}} = - \frac{1}{V}\frac{{dn}}{{dt}} = - \frac{1}{{RT}}\frac{{dP}}{{dt}}

) is correct because it correctly relates the rate of change of concentration, moles, and pressure using the ideal gas law.

Solution approach

Correct Answer + Core Reason: $- \frac{{dc}}{{dt}} = - \frac{1}{V}\frac{{dn}}{{dt}} = - \frac{1}{{RT}}\frac{{dP}}{{dt}}$ . This equation correctly relates the rate of change of concentration, moles, and pressure.

Step-by-Step Solution:

Step 1: Recall the ideal gas law: $PV = nRT$ .

Step 2: Concentration $c = \frac{n}{V}$ , so $n = cV$ .

Step 3: Differentiating with respect to time, $\frac{dn}{dt} = V \frac{dc}{dt}$ . Therefore, $\frac{dc}{dt} = \frac{1}{V} \frac{dn}{dt}$ .

Step 4: From the ideal gas law, $P = \frac{n}{V} RT = cRT$ . Differentiating with respect to time, $\frac{dP}{dt} = RT \frac{dc}{dt}$ . Therefore, $\frac{dc}{dt} = \frac{1}{RT} \frac{dP}{dt}$ .

Step 5: Since the reactant is being consumed, we use the negative sign to represent the rate of disappearance. Therefore, $- \frac{{dc}}{{dt}} = - \frac{1}{V}\frac{{dn}}{{dt}} = - \frac{1}{{RT}}\frac{{dP}}{{dt}}$ .

Why other options are wrong:
- Option B: $\frac{{dc}}{{dt}} = \frac{{dn}}{{dt}} = - \frac{{dP}}{{dt}}$ : This is incorrect because it does not account for the volume and temperature factors.
- Option C: $\frac{{dc}}{{dt}} = \frac{{RT}}{V}\frac{{dn}}{{dt}} = - \frac{{dP}}{{dt}}$ : This is incorrect because it has the wrong relationship between concentration and moles.
- Option D: All of the above: Only option A is correct.

Key Concepts:
- Ideal gas law
- Rate of reaction
- Relationship between concentration, moles, and pressure

Memory Aids:
- Remember the ideal gas law: $PV = nRT$ .

Quick Tips:
- Use the ideal gas law to relate pressure, volume, moles, and temperature.

4 Medium

4 of 71Question permalink

Analyzing +4 | -1

For a reaction of the type

A + B \to x

products, it is observed that doubling the concentration of

A

causes the reaction rate

({k_1})

to be four times as great but doubling the amount of

B

does not affect the rate

({k_2})

. The rate equation is

View answer and explanation

Correct option: B

k = {k_1}{k_2}

Option B (

k = {k_1}{k_2}

) is correct because the rate constant is affected by both A and B, and their effects are multiplicative.

Solution approach

Correct Answer + Core Reason: $k = {k_1}{k_2}$ . The rate constant is affected by both A and B, and their effects are multiplicative.

Step-by-Step Solution:

Step 1: Write the general rate law: Rate = $k[A]^m[B]^n$ , where m and n are the orders with respect to A and B, respectively.

Step 2: Analyze the effect of changing [A]: Doubling [A] causes the rate to increase by a factor of 4. Therefore, $4 = (2)^m$ , which means $m = 2$ .

Step 3: Analyze the effect of changing [B]: Doubling [B] does not affect the rate. Therefore, $n = 0$ .

Step 4: The rate law is Rate = $k[A]^2[B]^0 = k[A]^2$ .

Step 5: Since the rate is only dependent on A, $k_1$ represents the change in rate due to A, and $k_2$ represents the rate when B is changed. Since B has no effect, $k_2$ is essentially 1. Therefore, the overall rate constant is influenced by $k_1$ and $k_2$ multiplicatively, although $k_2$ is 1.

Step 6: While the problem is poorly worded, the best interpretation is that $k = k_1 * k_2$ where $k_2$ represents the factor by which the rate changes when [B] is changed. Since [B] has no effect, $k_2 = 1$ .

Why other options are wrong:
- Option A: This suggests that the rate constants are additive, which is not how rate constants combine in a rate law.
- Option C: This suggests that the rate constants are related by division, which is not consistent with the rate law.
- Option D: This option involves a square root and multiplication, which does not accurately represent the relationship between the rate constants and the orders of the reactants.

Key Concepts:
- Rate law
- Order of reaction
- Rate constant

Memory Aids:
- The rate law expresses the relationship between the rate of a reaction and the concentrations of the reactants.
- The order of a reaction with respect to a particular reactant is the exponent of that reactant's concentration in the rate law.

Quick Tips:
- Determine the order of the reaction with respect to each reactant by analyzing how the rate changes when the concentration of that reactant is changed.
- The overall order of the reaction is the sum of the orders with respect to each reactant.

5 Medium

5 of 71Question permalink

Analyzing +4 | -1

A hypothetical reaction

{A_2} + {B_2} \to 2AB

follows the mechanism as given below,

{A_2}\;\leftrightharpoons\;A + A\left( {fast} \right)

A + {B_2} \to AB + B(slow)

A + B \to AB\left( {fast} \right)

The order of the overall reaction is

View answer and explanation

Correct option: C

1\frac{1}{2}

Option C (

1\frac{1}{2}

) is correct because the rate law is determined by the slow step and the equilibrium of the fast step.

Solution approach

Correct Answer + Core Reason: $1\frac{1}{2}$ . The order of the reaction is determined from the rate law derived from the slow step and the equilibrium of the fast step.

Step-by-Step Solution:

Step 1: Identify the slow step: $A + {B_2} \to AB + B$ (slow).

Step 2: Write the rate law for the slow step: $rate = k[A][B_2]$ .

Step 3: Consider the fast equilibrium step: ${A_2} \rightleftharpoons A + A$ (fast).

Step 4: Write the equilibrium constant expression for the fast step: $K = \frac{{[A]^2}}{{[A_2]}}$ .

Step 5: Solve for [A]: $[A] = \sqrt{K[A_2]}$ .

Step 6: Substitute [A] into the rate law: $rate = k\sqrt{K[A_2]}[B_2] = k' [A_2]^{1/2} [B_2]$ .

Step 7: Determine the overall order of the reaction: Order = $\frac{1}{2} + 1 = \frac{3}{2} = 1\frac{1}{2}$ .

Why other options are wrong:
- Option A: 2: This is incorrect. It might be obtained by not considering the equilibrium of the fast step.
- Option B: 1: This is incorrect. It may result from only considering the slow step without accounting for the fast equilibrium.
- Option D: 0: This is incorrect. It may be a calculation error during the process.

Key Concepts:
- Rate Law
- Order of Reaction
- Reaction Mechanism
- Equilibrium

Memory Aids:
- The rate law is determined by the slow step.

Quick Tips:
- Identify the slow step and the fast equilibrium steps.
- Express the concentration of intermediates in terms of reactants using the equilibrium constant.

6 Medium

6 of 71Question permalink

Application +4 | -1

The activation energy of exothermic reaction

A \to B

is

80\;kJ\;\mathrm{mo}{\mathrm{l}^{ - 1}}

. The heat of reaction is

200\;kJ\;\;\mathrm{mo}{\mathrm{l}^{ - 1}}

. The activation energy for the reaction

B \to A

(in

kJ\;\;\mathrm{mo}{\mathrm{l}^{ - 1}})

will be

View answer and explanation

Correct option: D

280

Option D (280) is correct because the activation energy for the reverse reaction is the sum of the activation energy for the forward reaction and the heat of reaction.

Solution approach

Correct Answer + Core Reason: 280. The activation energy for the reverse reaction is the sum of the activation energy for the forward reaction and the heat of reaction.

Step-by-Step Solution:

Step 1: Understand the relationship between activation energies and heat of reaction: $E_{a(reverse)} = E_{a(forward)} + |\Delta H|$ . Since the reaction is exothermic, $\Delta H$ is negative.

Step 2: Substitute the given values: $E_{a(reverse)} = 80 + 200 = 280 \; kJ \; mol^{-1}$ .

Why other options are wrong:
- Option A: 80: This is the activation energy for the forward reaction.
- Option B: 120: This is incorrect.
- Option C: 40: This is incorrect.

Key Concepts:
- Activation Energy
- Heat of Reaction
- Exothermic Reactions

Memory Aids:
- Remember the relationship between activation energies and heat of reaction.

Quick Tips:
- Pay attention to the sign of the heat of reaction.

7 Medium

7 of 71Question permalink

Application +4 | -1

For a reaction

A + B\, \to

Products, the rate of the reaction was doubled when the concentration of

A

was doubled. When the concentration of

A

and

B

were doubled, the rate was again doubled, the order of the reaction with respect to

A

and

B

are:

View answer and explanation

Correct option: C

1,0

Option C (1,0) is correct because doubling the concentration of A doubles the rate, indicating a first-order dependence on A, while changing B has no effect.

Solution approach

Correct Answer + Core Reason: 1,0. The reaction is first order with respect to A and zero order with respect to B.

Step-by-Step Solution:

Step 1: Let the rate law be: Rate = $k[A]^x[B]^y$ , where x is the order with respect to A and y is the order with respect to B.

Step 2: When the concentration of A is doubled, the rate doubles. So, if $[A]_2 = 2[A]_1$ , then Rate $_2 = 2$ Rate $_1$ .

Step 3: When the concentrations of A and B are doubled, the rate doubles. So, if $[A]_3 = 2[A]_1$ and $[B]_3 = 2[B]_1$ , then Rate $_3 = 2$ Rate $_1$ .

Step 4: Write the rate laws for the three conditions:\nRate $_1 = k[A]_1^x[B]_1^y$ \nRate $_2 = k[2A]_1^x[B]_1^y = 2$ Rate $_1$ \nRate $_3 = k[2A]_1^x[2B]_1^y = 2$ Rate $_1$

Step 5: From Rate $_2 = 2$ Rate $_1$ , we have $k[2A]_1^x[B]_1^y = 2k[A]_1^x[B]_1^y$ . This simplifies to $2^x = 2$ , so $x = 1$ .

Step 6: From Rate $_3 = 2$ Rate $_1$ , we have $k[2A]_1^x[2B]_1^y = 2k[A]_1^x[B]_1^y$ . Since we know $x = 1$ , we have $k[2A]_1[2B]_1^y = 2k[A]_1[B]_1^y$ . This simplifies to $2[2B]_1^y = 2[B]_1^y$ , so $[2B]_1^y = [B]_1^y$ . This means $y = 0$ .

Why other options are wrong:
- Option A: This option is incorrect because if the order was 1,1, doubling both A and B would quadruple the rate.
- Option B: This option is incorrect because if the order was 2,0, doubling A would quadruple the rate.
- Option D: This option is incorrect because if the order was 0,1, doubling B would double the rate, but doubling A has no effect.

Key Concepts:
- Rate law and order of reaction

Memory Aids:
- Remember that the order of the reaction determines how the rate changes with concentration.

Quick Tips:
- Use the given information to set up a system of equations to solve for the orders of the reaction.

8 Medium

8 of 71Question permalink

Remembering +4 | -1

A catalyst alters, which of the following in a chemical reaction?

View answer and explanation

Correct option: D

Activation energy

Option D is correct because a catalyst lowers the activation energy of a reaction, providing an alternative pathway with a lower energy barrier.

Solution approach

Correct Answer + Core Reason: Activation energy. A catalyst alters the activation energy of a chemical reaction.

Step-by-Step Solution:

Step 1: Define catalyst. A catalyst is a substance that speeds up a chemical reaction without being consumed in the reaction.

Step 2: Explain how a catalyst works. Catalysts lower the activation energy ( $E_a$ ) of a reaction by providing an alternative reaction pathway with a lower energy barrier. This allows more molecules to have sufficient energy to react at a given temperature.

Why other options are wrong:
- Option A: Entropy is a measure of disorder and is not directly altered by a catalyst.
- Option B: Enthalpy is the heat content of a system and is not directly altered by a catalyst. The catalyst only affects the rate, not the thermodynamics.
- Option C: Internal energy is the total energy of a system and is not directly altered by a catalyst.

Key Concepts:
- Catalyst
- Activation energy
- Reaction mechanism

Memory Aids:
- Catalyst lowers activation energy.

Quick Tips:
- Catalysts do not change the equilibrium constant or the thermodynamics of a reaction, only the rate.

9 Medium

9 of 71Question permalink

Analyzing +4 | -1

Consider the reaction,

C{l_2}\left( {aq} \right) + {H_2}S\left( {aq} \right) \to S\left( s \right) + 2{H^ + }\left( {aq} \right) + 2C{l^ - }\left( {aq} \right)

the rate equation for this reaction is ,

rate = k\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]

Which of these mechanisms is/are consistent with this rate equation?

\begin{array}{l} A.\;\;C{l_2} + {H_2}S\left( {aq} \right) \to {H^ + }C{l^ - } + C{l^ + } + H{S^ - }\left( {slow} \right)\\ \;\;\;\;\;C{l^ + } + H{S^ - } \to {H^ + }C{l^ - } + S\left( {fast} \right)\\ B.\;\;{H_2}S \leftrightarrow {H^ + } + H{S^ - }\left( {fast\;equilibrium} \right)\\ \;\;\;\;C{l_2} + H{S^ - } \to 2C{l^ - } + {H^ + }S\left( {slow} \right) \end{array}

View answer and explanation

Correct option: D

(A)Only

Option D (All of the above) is correct because all the statements about the rate constants and equilibrium constant are derived correctly from the given net rate equation.

Solution approach

Correct Answer + Core Reason: All of the above. All the statements are correct based on the given net rate equation.

Step-by-Step Solution:

Step 1: Analyze the given net rate equation:\nnet $rate = 2.0 \times {10^{ - 4}}\;{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}\left[ {{\mathrm{H}_2}\mathrm{O}} \right] - 3.0 \times {10^5}{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}{\mathrm{H}_2}\mathrm{O}} \right]^{2 + }} \cdot \left[ {\mathrm{N}{\mathrm{H}_3}} \right]$

Step 2: Identify the forward and backward rate constants:
- Forward rate constant: ${k_f} = 2.0 \times {10^{ - 4}}$
- Backward rate constant: ${k_b} = 3.0 \times {10^5}$

Step 3: Calculate the equilibrium constant:
$K = \frac{{{k_f}}}{{{k_b}}} = \frac{{2.0 \times {{10}^{ - 4}}}}{{3.0 \times {{10}^5}}} = 6.67 \times {10^{ - 10}} \approx 6.6 \times {10^{ - 10}}$

Step 4: Evaluate each statement:
- Option A: Rate constant for forward reaction $= 2 \times {10^{ - 4}}$ . This is correct.
- Option B: Rate constant for backward reaction $= 3 \times {10^5}$ . This is correct.
- Option C: Equilibrium constant for the reaction $= 6.6 \times {10^{ - 10}}$ . This is correct.

Step 5: Since all statements are correct, the answer is All of the above.

Why other options are wrong:
- Option A: This is only one of the correct statements.
- Option B: This is only one of the correct statements.
- Option C: This is only one of the correct statements.

Key Concepts:
- Rate law
- Rate constant
- Equilibrium constant

Memory Aids:
- Remember that $K = \frac{{{k_f}}}{{{k_b}}}$ .

Quick Tips:
- Identify the forward and backward rate constants from the net rate equation.

10 Medium

10 of 71Question permalink

Analyzing +4 | -1

The reaction obey I order with respect to

{\mathrm{H}_2}

and

\mathrm{ICl}

both

{\mathrm{H}_2}\left( \mathrm{g} \right) + 2\mathrm{ICl}\left( \mathrm{g} \right) \to 2\mathrm{HCl}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)

Which of the following mechanism is in consistent with the given fact?

Mechanism

A:{\mathrm{H}_2}\left( \mathrm{g} \right) + 2\mathrm{ICl} \to 2\mathrm{HCl}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)

Mechanism

B:\left( \mathrm{i} \right){\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{ICl}\left( \mathrm{g} \right)\xrightarrow{{Slow}}\mathrm{HCl}\left( \mathrm{g} \right) + \mathrm{HI}\left( \mathrm{g} \right)

\left( {\mathrm{ii}} \right)\mathrm{HI}\left( \mathrm{g} \right) + \mathrm{ICl}\left( \mathrm{g} \right) \to \mathrm{HCl}\left( \mathrm{g} \right) + {\mathrm{I}_2}

View answer and explanation

Correct option: D

B

only

Option D (

B

only) is correct because Mechanism B is consistent with the rate law, while Mechanism A is not.

Solution approach

Correct Answer + Core Reason: $B$ only. Mechanism B is consistent with the given rate law, while Mechanism A implies a different rate law.

Step-by-Step Solution:

Step 1: Write the given rate law: Rate = $k[H_2][ICl]$

Step 2: Analyze Mechanism A:
- Mechanism A: ${\mathrm{H}_2}\left( \mathrm{g} \right) + 2\mathrm{ICl} \to 2\mathrm{HCl}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)$ . If this is an elementary reaction, the rate law would be Rate = $k[H_2][ICl]^2$ , which is inconsistent with the given rate law.

Step 3: Analyze Mechanism B:
- (i) ${\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{ICl}\left( \mathrm{g} \right)\xrightarrow{{Slow}}\mathrm{HCl}\left( \mathrm{g} \right) + \mathrm{HI}\left( \mathrm{g} \right)$ . The rate law for this step is Rate = $k[H_2][ICl]$ .
- (ii) $\mathrm{HI}\left( \mathrm{g} \right) + \mathrm{ICl}\left( \mathrm{g} \right) \to \mathrm{HCl}\left( \mathrm{g} \right) + {\mathrm{I}_2}$ . This step is fast and does not affect the overall rate.
- The overall rate is determined by the slow step, which is Rate = $k[H_2][ICl]$ , consistent with the given rate law.

Step 4: Therefore, Mechanism B is consistent with the given rate law, while Mechanism A is not.

Why other options are wrong:
- Option A: $A$ and $B$ both: Mechanism A is inconsistent with the given rate law.
- Option B: Neither $A$ nor $B$ : Mechanism B is consistent with the given rate law.
- Option C: $A$ only: Mechanism A is inconsistent with the given rate law.

Key Concepts:
- Rate law
- Reaction mechanism
- Rate-determining step

Memory Aids:
- The rate law is determined by the slow step in a reaction mechanism.

Quick Tips:
- Identify the slow step in the mechanism and write its rate law. Compare it with the given rate law.

11 Medium

11 of 71Question permalink

Analyzing +4 | -1

Consider the following statements,

The rate law for the acid catalysed hydrolysis of an ester being given as

Rate = k\;\left[ {{H^ + }} \right]\;\left[ {ester} \right] = k'\;\left[ {ester} \right]

If the acid concentration is doubled at constant ester concentration

1. The second order rate constant,

k

is doubled.

2. The pseudo first order rate constant,

k

is double.

3. The rate of the reaction is doubled.

Which of the above statements are correct?

View answer and explanation

Correct option: B

2 and 3

Option B (2 and 3) is correct because doubling the acid concentration doubles the pseudo-first-order rate constant and the overall reaction rate, while the second-order rate constant remains unchanged.

Solution approach

Correct Answer + Core Reason: 2 and 3. Doubling the acid concentration doubles the pseudo-first-order rate constant and the rate of the reaction.

Step-by-Step Solution:

Step 1: Understand the rate law: $Rate = k[H^+][ester] = k'[ester]$

Step 2: Identify $k'$ as the pseudo-first-order rate constant: $k' = k[H^+]$

Step 3: Analyze the effect of doubling $[H^+]$ :\nIf $[H^+]$ is doubled, $k'$ becomes $k(2[H^+]) = 2k[H^+] = 2k'$ \nSo, $k'$ is doubled.

Step 4: Analyze the effect on the rate:\nIf $[H^+]$ is doubled, the new rate is $Rate_{new} = k(2[H^+][ester]) = 2k[H^+][ester] = 2 \times Rate$ \nSo, the rate is doubled.

Step 5: The second-order rate constant, $k$ , remains unchanged as it's a constant for the reaction at a given temperature.

Why other options are wrong:
- Option A (1 and 2): The second-order rate constant $k$ does not change with concentration.
- Option C (1 and 3): The second-order rate constant $k$ does not change with concentration.
- Option D (1, 2 and 3): The second-order rate constant $k$ does not change with concentration.

Key Concepts:
- Rate Laws
- Pseudo-First-Order Reactions
- Effect of Concentration on Reaction Rate

Memory Aids:
- Remember that rate constants are temperature-dependent, not concentration-dependent.

Quick Tips:
- Distinguish between the overall rate constant and the pseudo-first-order rate constant.

12 Medium

12 of 71Question permalink

Application +4 | -1

The activation energies of forward and backward reaction:

{A_2} + {B_2}\leftrightharpoons2AB

are 180kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

and 200 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

respectively. The presence of a catalyst lowers the activation energy of both (forward and backward) reactions by 100 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

. The enthalpy change of the reaction in the presence of catalyst will be (in kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

):

View answer and explanation

Correct option: A

- 20

Option A (

- 20

) is correct because the enthalpy change is the difference between the activation energies of the forward and backward reactions, and the catalyst lowers both by the same amount, so the enthalpy change remains the same.

Solution approach

Correct Answer + Core Reason: $- 20$ . The enthalpy change is the difference between the activation energies of the forward and backward reactions, and a catalyst lowers both by the same amount.

Step-by-Step Solution:

Step 1: Calculate the enthalpy change without the catalyst: $\Delta H = E_{a(forward)} - E_{a(backward)} = 180 - 200 = -20 \; kJ \; mol^{-1}$ .

Step 2: Calculate the activation energies with the catalyst: $E'_{a(forward)} = 180 - 100 = 80 \; kJ \; mol^{-1}$ , and $E'_{a(backward)} = 200 - 100 = 100 \; kJ \; mol^{-1}$ .

Step 3: Calculate the enthalpy change with the catalyst: $\Delta H' = E'_{a(forward)} - E'_{a(backward)} = 80 - 100 = -20 \; kJ \; mol^{-1}$ .

Why other options are wrong:
- Option B: $- 300$ : This is incorrect.
- Option C: $+ 120$ : This is incorrect.
- Option D: $- 280$ : This is incorrect.

Key Concepts:
- Activation Energy
- Enthalpy Change
- Catalyst

Memory Aids:
- A catalyst lowers the activation energy but does not change the enthalpy change.

Quick Tips:
- The enthalpy change is independent of the catalyst.

13 Medium

13 of 71Question permalink

Understanding +4 | -1

The rate for the reaction,

RCl + NaOH\left( {aq} \right) \to ROH + NaCl

is given by

rate\mathrm{ } = K_1\left[ {RCl} \right].

The rate of the reaction is

View answer and explanation

Correct option: C

Halved on reducing the concentration of

RCl

to half

Option C (Halved on reducing the concentration of

RCl

to half) is correct because the rate is directly proportional to the concentration of

RCl

in a first-order reaction.

Solution approach

Correct Answer + Core Reason: Halved on reducing the concentration of $RCl$ to half. The rate is directly proportional to the concentration of $RCl$ .

Step-by-Step Solution:

Step 1: The rate equation is given as $rate = K_1[RCl]$ .

Step 2: This is a first-order reaction with respect to $RCl$ .

Step 3: If the concentration of $RCl$ is reduced to half, the rate of the reaction will also be halved.

Step 4: The rate is independent of the concentration of $NaOH$ since it does not appear in the rate equation.

Step 5: Increasing the temperature will increase the rate constant $K_1$ , and therefore increase the rate of the reaction.

Why other options are wrong:
- Option A: Unaffected by increasing the temperature of the reaction: Increasing the temperature will increase the rate of the reaction.
- Option B: Decreased on increasing the temperature of the reaction: Increasing the temperature will increase the rate of the reaction.
- Option D: Doubled on doubling the concentration of $NaOH$ : The rate is independent of the concentration of $NaOH$ .

Key Concepts:
- Rate law
- Order of reaction

Memory Aids:
- The rate is directly proportional to the concentration of the reactant in a first-order reaction.

Quick Tips:
- Understand how the rate depends on the concentration of reactants based on the rate law.

14 Medium

14 of 71Question permalink

Application +4 | -1

A first order reaction is 20% complete in 10 min. What is the rate constant of the reaction?

View answer and explanation

Correct option: B

0.0223

Option B (0.0223) is correct because it is the rate constant calculated using the integrated rate law for a first-order reaction with 20% completion in 10 minutes.

Solution approach

Correct Answer + Core Reason: 0.0223. The rate constant is calculated using the integrated rate law for a first-order reaction.

Step-by-Step Solution:

Step 1: Identify the given information: 20% completion in 10 minutes. This means that 80% of the reactant remains.

Step 2: Recall the integrated rate law for a first-order reaction: $k = \frac{{2.303}}{t}\log \frac{{[A]_0}}{{[A]_t}}$ .

Step 3: Let $[A]_0 = 100$ and $[A]_t = 80$ . Then, $k = \frac{{2.303}}{{10}}\log \frac{{100}}{{80}}$ .

Step 4: Simplify the expression: $k = \frac{{2.303}}{{10}}\log 1.25 = \frac{{2.303}}{{10}} \times 0.0969 = 0.0223$ .

Why other options are wrong:
- Option A: This option is incorrect as it involves an incorrect calculation using the given values.
- Option C: This option is incorrect as it involves an incorrect calculation using the given values.
- Option D: This option is incorrect as it involves an incorrect calculation using the given values.

Key Concepts:
- Integrated rate law for first-order reactions

Memory Aids:
- Remember the integrated rate law and how to use the percentage of completion to find the remaining concentration.

Quick Tips:
- Always use the integrated rate law when dealing with time and concentrations in reaction kinetics.

15 Medium

15 of 71Question permalink

Understanding +4 | -1

At room temperature, the reaction between

\mathrm{NO}

and

{\mathrm{O}_2}

to give

\mathrm{N}{\mathrm{O}_2}

is fast, while that between

\mathrm{CO}

and

{\mathrm{O}_2}

is slow. It is due to:

View answer and explanation

Correct option: C

The activation energy for the reaction,

2\mathrm{NO} + {\mathrm{O}_2} \to 2\mathrm{N}{\mathrm{O}_2}

is less than

2\mathrm{CO} + {\mathrm{O}_2} \to 2\mathrm{C}{\mathrm{O}_2}

Option C (The activation energy for the reaction,

2\mathrm{NO} + {\mathrm{O}_2} \to 2\mathrm{N}{\mathrm{O}_2}

is less than

2\mathrm{CO} + {\mathrm{O}_2} \to 2\mathrm{C}{\mathrm{O}_2}

) is correct because reactions with lower activation energies proceed faster.

Solution approach

Correct Answer + Core Reason: The activation energy for the reaction, $2\mathrm{NO} + {\mathrm{O}_2} \to 2\mathrm{N}{\mathrm{O}_2}$ is less than $2\mathrm{CO} + {\mathrm{O}_2} \to 2\mathrm{C}{\mathrm{O}_2}$ . Reactions with lower activation energies are faster.

Step-by-Step Solution:

Step 1: The rate of a reaction is inversely proportional to the activation energy.

Step 2: The reaction between $\mathrm{NO}$ and ${\mathrm{O}_2}$ is fast, while the reaction between $\mathrm{CO}$ and ${\mathrm{O}_2}$ is slow.

Step 3: This implies that the activation energy for the reaction between $\mathrm{NO}$ and ${\mathrm{O}_2}$ is less than the activation energy for the reaction between $\mathrm{CO}$ and ${\mathrm{O}_2}$ .

Why other options are wrong:
- Option A: $\mathrm{CO}$ is smaller in size than that of $\mathrm{NO}$ : Size difference does not directly explain the difference in reaction rates.
- Option B: $\mathrm{CO}$ is poisonous: Being poisonous is irrelevant to the reaction rate.
- Option D: None of the above: Option C is the correct explanation.

Key Concepts:
- Activation energy
- Reaction rate

Memory Aids:
- Lower activation energy means faster reaction.

Quick Tips:
- Consider the activation energy when comparing the rates of different reactions.

16 Medium

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Understanding +4 | -1

For a reaction of order

n

, the unit of the rate constant is

View answer and explanation

Correct option: C

\mathrm{mo}{\mathrm{l}^{1 - n}}{\mathrm{L}^{n - 1}}{\mathrm{s}^{ - 1}}

Option C (

\mathrm{mo}{\mathrm{l}^{1 - n}}{\mathrm{L}^{n - 1}}{\mathrm{s}^{ - 1}}

) is correct because this is the general formula for the units of the rate constant for a reaction of order n.

Solution approach

Correct Answer + Core Reason: $\mathrm{mo}{\mathrm{l}^{1 - n}}{\mathrm{L}^{n - 1}}{\mathrm{s}^{ - 1}}$ . This is the general formula for the units of the rate constant.

Step-by-Step Solution:

Step 1: Recall the general rate law: Rate = $k{\left[ A \right]^n}$ , where n is the order of the reaction.

Step 2: The units of rate are typically $mol \, L^{-1} s^{-1}$ .

Step 3: The units of concentration are $mol \, L^{-1}$ .

Step 4: Therefore, the units of k are: Units of Rate / (Units of Concentration)^n = $(mol \, L^{-1} s^{-1}) / (mol \, L^{-1})^n = mol^{1-n} L^{n-1} s^{-1}$

Why other options are wrong:
- Option A: Incorrect exponents for L.
- Option B: Incorrect exponents for L.
- Option D: Incorrect exponents for L.

Key Concepts:
- Rate constant
- Order of reaction
- Units of rate constant

Memory Aids:
- Remember the general formula for the units of the rate constant.

Quick Tips:
- Derive the units of the rate constant from the rate law.

17 Medium

17 of 71Question permalink

Understanding +4 | -1

What is the formula to find value of

{t_{1/2}}

for a zero order reaction?

View answer and explanation

Correct option: C

\frac{{{{\left[ R \right]}_0}}}{{2k}}

Option C (

\frac{{{{\left[ R \right]}_0}}}{{2k}}

) is correct because this is the standard formula for the half-life of a zero-order reaction.

Solution approach

Correct Answer + Core Reason: $\frac{{{{\left[ R \right]}_0}}}{{2k}}$ . This is the standard formula for the half-life of a zero-order reaction.

Step-by-Step Solution:

Step 1: Recall the rate law for a zero-order reaction: Rate = k

Step 2: Recall the integrated rate law for a zero-order reaction: $\left[ R \right] = {\left[ R \right]_0} - kt$

Step 3: Define half-life: The time it takes for the concentration of the reactant to decrease to half of its initial concentration, i.e., $\left[ R \right] = \frac{{{{\left[ R \right]}_0}}}{2}$

Step 4: Substitute $\left[ R \right] = \frac{{{{\left[ R \right]}_0}}}{2}$ into the integrated rate law: $\frac{{{{\left[ R \right]}_0}}}{2} = {\left[ R \right]_0} - k{t_{1/2}}$

Step 5: Solve for ${t_{1/2}}$ : $k{t_{1/2}} = {\left[ R \right]_0} - \frac{{{{\left[ R \right]}_0}}}{2} = \frac{{{{\left[ R \right]}_0}}}{2}$

Step 6: ${t_{1/2}} = \frac{{{{\left[ R \right]}_0}}}{{2k}}$

Why other options are wrong:
- Option A: This is the inverse of the correct formula and dimensionally incorrect.
- Option B: This is incorrect and dimensionally incorrect.
- Option D: This is the half-life formula for a first-order reaction, not a zero-order reaction.

Key Concepts:
- Zero-order reaction
- Half-life
- Integrated rate law

Memory Aids:
- Remember the zero-order half-life formula.

Quick Tips:
- Understand the derivation of the half-life formula from the integrated rate law.

18 Medium

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Remembering +4 | -1

Which of the following does not affect the rate of reaction?

View answer and explanation

Correct option: C

\Delta H

of reaction

Option C is correct because the enthalpy change (

\Delta H

) of a reaction is a thermodynamic property and does not directly influence the reaction rate.

Solution approach

Correct Answer + Core Reason: $\Delta H$ of reaction. The enthalpy change of a reaction does not directly affect the rate of the reaction.

Step-by-Step Solution:

Step 1: Identify factors affecting reaction rate. The rate of a reaction is affected by factors such as concentration of reactants, physical state of reactants, temperature, and presence of a catalyst.

Step 2: Explain why $\Delta H$ is not a direct factor. $\Delta H$ is a thermodynamic property that indicates whether a reaction is endothermic or exothermic. It does not directly influence the kinetics (rate) of the reaction.

Why other options are wrong:
- Option A: Amount of the reactant taken affects the concentration, which affects the rate.
- Option B: Physical state of the reactant affects the rate because it influences the frequency of collisions between reactant molecules.
- Option D: Size of vessel can affect the concentration of reactants, thus affecting the rate.

Key Concepts:
- Rate of reaction
- Enthalpy change
- Factors affecting reaction rate

Memory Aids:
- Rate is kinetics, enthalpy is thermodynamics.

Quick Tips:
- Focus on factors that directly influence the frequency and energy of collisions between reactant molecules.

19 Medium

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Understanding +4 | -1

Which statement is correct?

View answer and explanation

Correct option: D

Increase in the total pressure of a gas phase reaction increase the fraction of collisions effective in producing reactions

Option D (Increase in the total pressure of a gas phase reaction increase the fraction of collisions effective in producing reactions) is correct because increasing pressure increases the collision frequency and the likelihood of effective collisions.

Solution approach

Correct Answer + Core Reason: Increase in the total pressure of a gas phase reaction increase the fraction of collisions effective in producing reactions. Increasing pressure increases collision frequency and effective collisions.

Step-by-Step Solution:

Step 1: Analyze each statement:
- A: Reactions with low activation energy are usually exothermic: This is not always true. Activation energy and enthalpy change are independent.
- B: The rate law sometimes enable to deduce the mechanism of a reaction: This is true, but not always. Rate law can provide clues about the mechanism.
- C: The rate law for reaction is an algebraic expression relating the forward reaction rate to product concentration: The rate law relates the rate to reactant concentrations, not product concentrations.
- D: Increase in the total pressure of a gas phase reaction increase the fraction of collisions effective in producing reactions: Increasing pressure increases the concentration of reactants, leading to more frequent and effective collisions.

Step 2: Based on the analysis, statement D is the most accurate.

Why other options are wrong:
- Option A: Reactions with low activation energy are usually exothermic: Activation energy and enthalpy change are independent.
- Option B: The rate law sometimes enable to deduce the mechanism of a reaction: While rate law can provide clues, it doesn't always allow deduction of the mechanism.
- Option C: The rate law for reaction is an algebraic expression relating the forward reaction rate to product concentration: The rate law relates the rate to reactant concentrations, not product concentrations.

Key Concepts:
- Activation energy
- Rate law
- Reaction mechanism
- Collision theory

Memory Aids:
- Remember that increasing pressure increases collision frequency.

Quick Tips:
- Carefully analyze each statement and consider the underlying principles.

20 Medium

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Understanding +4 | -1

Which one is not correct?

View answer and explanation

Correct option: A

Rate of zero order reaction depends upon initial concentration of reactant

Option A (Rate of zero order reaction depends upon initial concentration of reactant) is correct because the rate of a zero-order reaction is independent of the concentration of the reactant.

Solution approach

Correct Answer + Core Reason: Rate of zero order reaction depends upon initial concentration of reactant. The rate of a zero-order reaction is independent of the concentration of the reactant.

Step-by-Step Solution:

Step 1: Define a zero-order reaction: A reaction whose rate is independent of the concentration of the reactant(s).

Step 2: The rate law for a zero-order reaction is: rate = $k$ , where $k$ is the rate constant.

Step 3: This rate law shows that the rate is constant and does not depend on the concentration of the reactant.

Why other options are wrong:
- Option B: Rate of zero order reaction does not depend upon initial concentration of reactant: This statement is correct.
- Option C: ${t_{1/2}}$ of first order reaction is independent of initial concentration of reaction: This statement is correct.
- Option D: ${t_{1/2}}$ of zero order reaction is dependent of initial concentration of reaction: This statement is correct.

Key Concepts:
- Zero-order reaction
- Rate law
- Half-life

Memory Aids:
- Zero-order: rate = k

Quick Tips:
- Remember the rate laws for different orders of reactions.

21 Medium

21 of 71Question permalink

Remembering +4 | -1

Rate constant for a reaction is

\lambda

. Average life is represent by

View answer and explanation

Correct option: A

\frac{1}{\lambda }

Option A (

\frac{1}{\lambda }

) is correct because the average life is the reciprocal of the rate constant for a first-order reaction.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\lambda }$ . The average life is the reciprocal of the rate constant.

Step-by-Step Solution:

Step 1: Define average life ( $\tau$ ). The average life of a reactant is the average time for which a molecule exists before it reacts.

Step 2: Relate average life to the rate constant ( $\lambda$ ) for a first-order reaction. The average life is the reciprocal of the rate constant: $\tau = \frac{1}{\lambda}$ .

Step 3: Conclusion. Therefore, the average life is $\frac{1}{\lambda}$ .

Why other options are wrong:
- **Option B ( $\frac{{1{n^2}}}{\lambda }$ ): This is not the correct formula for average life.
- Option C ( $\frac{\lambda }{{\surd 2}}$ ): This is not the correct formula for average life.
- Option D ( $\frac{{0.693}}{\lambda }$ ): This is the formula for half-life, not average life.

Key Concepts:
- Average life
- Rate constant
- First-order kinetics

Memory Aids:
- Remember that average life is the reciprocal of the rate constant.

Quick Tips:**
- Distinguish between average life and half-life.

22 Medium

22 of 71Question permalink

Application +4 | -1

The rate constant of a reaction at temperature

200\;K

is 10 times less than the rate constant at

400\;K

. What is the activation energy (

{E_a}

) of the reaction?

View answer and explanation

Correct option: B

921.2\;R

Option B (

921.2\;R

) is correct because it is the activation energy calculated using the Arrhenius equation, given the rate constants at two different temperatures.

Solution approach

Correct Answer + Core Reason: $921.2\;R$ . The activation energy is calculated using the Arrhenius equation and the given rate constants at two different temperatures.

Step-by-Step Solution:

Step 1: Write the Arrhenius equation: $k = A e^{-E_a/RT}$

Step 2: We are given that $k_1 = \frac{k_2}{10}$ at $T_1 = 200 K$ and $T_2 = 400 K$ .

Step 3: Use the equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$

Step 4: Substitute the given values: $\ln(10) = \frac{E_a}{R} (\frac{1}{200} - \frac{1}{400})$

Step 5: Simplify: $2.303 = \frac{E_a}{R} (\frac{2-1}{400}) = \frac{E_a}{R} (\frac{1}{400})$

Step 6: Solve for $E_a$ : $E_a = 2.303 \times 400 \times R = 921.2 R$

Step 7: Final Answer: $921.2\;R$

Why other options are wrong:
- Option A: This is incorrect because it's twice the correct value.
- Option C: This is incorrect because it's half the correct value.
- Option D: This is incorrect because it's a quarter of the correct value.

Key Concepts:
- Arrhenius equation
- Activation energy

Memory Aids:
- Remember the Arrhenius equation and how to use it to calculate activation energy.

Quick Tips:
- Pay attention to the units of the gas constant R and the temperatures.

23 Medium

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Analyzing +4 | -1

Which statement is not correct?

View answer and explanation

Correct option: B

For exothermic reactions, heat of reaction is more than energy of activation

Option B (For exothermic reactions, heat of reaction is more than energy of activation) is correct because for exothermic reactions, the heat of reaction is negative, and its magnitude is typically less than the activation energy.

Solution approach

Correct Answer + Core Reason: For exothermic reactions, heat of reaction is more than energy of activation. This statement is incorrect because the heat of reaction for exothermic reactions is typically less than the activation energy.

Step-by-Step Solution:

Step 1: Analyze each statement:
- Option A: For endothermic reactions, heat of reaction is lesser than energy of activation. This is generally correct. Endothermic reactions require energy input, so the activation energy is higher than the heat of reaction.
- Option B: For exothermic reactions, heat of reaction is more than energy of activation. This is incorrect. For exothermic reactions, the heat of reaction is negative, and its magnitude is typically less than the activation energy.
- Option C: For exothermic reactions energy of activation is less in forward reaction than in backward reaction. This is correct.
- Option D: For endothermic reactions energy of activation is more in forward reaction than in backward reaction. This is correct.

Step 2: Identify the incorrect statement. Option B is incorrect because for exothermic reactions, the heat of reaction is negative, and its magnitude is typically less than the activation energy.

Why other options are wrong:
- Option A: For endothermic reactions, heat of reaction is lesser than energy of activation: This is generally correct.
- Option C: For exothermic reactions energy of activation is less in forward reaction than in backward reaction: This is correct.
- Option D: For endothermic reactions energy of activation is more in forward reaction than in backward reaction: This is correct.

Key Concepts:
- Endothermic Reactions
- Exothermic Reactions
- Activation Energy
- Heat of Reaction

Memory Aids:
- Endothermic reactions absorb heat, exothermic reactions release heat.

Quick Tips:
- Draw energy diagrams to visualize the relationship between activation energy and heat of reaction.

24 Medium

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Analyzing +4 | -1

The data for the reaction,

\mathrm{ }A + B \to C

\begin{array}{l} Ex\;\;\;\;\;\;\;\;\;\;\;{\left[ A \right]_0}\;\;\;\;\;\;\;\;\;{\left[ B \right]_0}\;\;\;\;\;\;\;\;\;Initial\;\mathrm{ }rate\\ \begin{array}{llllllllllllllllllll} {1\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.012\;\;\;\;\;\;\;\;0.035\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.10}\\ {2\;\;\;\;\;\;\;\;\;\;\;\;\;0.024\;\;\;\;\;\;\;\;\;0.070\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.80}\\ {3\;\;\;\;\;\;\;\;\;\;\;\;\;0.024\;\;\;\;\;\;\;\;\;\;0.035\;\;\;\;\;\;\;\;\;\;\;\;\;0.10}\\ {4\;\;\;\;\;\;\;\;\;\;\;\;\;0.012\;\;\;\;\;\;\;\;\;0.070\;\;\;\;\;\;\;\;\;\;\;\;\;0.80} \end{array} \end{array}

The rate law corresponds to the above data is

View answer and explanation

Correct option: A

rate = k{\left[ B \right]^3}

Option A (

rate = k{\left[ B \right]^3}

) is correct because the rate is only dependent on the concentration of B, and the order with respect to B is 3.

Solution approach

Correct Answer + Core Reason: $rate = k{\left[ B \right]^3}$ . The rate law is determined by analyzing the given experimental data.

Step-by-Step Solution:

Step 1: Analyze the given data:
- Experiment 1: $[A]_0 = 0.012$ , $[B]_0 = 0.035$ , Initial rate = 0.10
- Experiment 2: $[A]_0 = 0.024$ , $[B]_0 = 0.070$ , Initial rate = 0.80
- Experiment 3: $[A]_0 = 0.024$ , $[B]_0 = 0.035$ , Initial rate = 0.10
- Experiment 4: $[A]_0 = 0.012$ , $[B]_0 = 0.070$ , Initial rate = 0.80

Step 2: Compare experiments 1 and 3. When [A] doubles and [B] is constant, the rate remains constant. This means the reaction is zero order with respect to A.

Step 3: Compare experiments 1 and 2. When [A] doubles and [B] doubles, the rate increases by a factor of 8. Since the reaction is zero order with respect to A, the rate increase is only due to the change in [B].

Step 4: Let the rate law be $rate = k[A]^x[B]^y$ . Since the reaction is zero order with respect to A, $x = 0$ .\nSo, $rate = k[B]^y$ .

Step 5: From experiments 1 and 2:
$\frac{rate_2}{rate_1} = \frac{k[B_2]^y}{k[B_1]^y} = \frac{0.80}{0.10} = 8$
$\frac{[0.070]^y}{[0.035]^y} = 8$
$(2)^y = 8$
$y = 3$

Step 6: Therefore, the rate law is $rate = k[B]^3$ .

Why other options are wrong:
- Option B: $rate = k{\left[ B \right]^4}$ : This is incorrect. The order with respect to B is 3, not 4.
- Option C: $rate = k\left[ A \right]{\left[ B \right]^3}$ : This is incorrect. The reaction is zero order with respect to A.
- Option D: $rate = k{\left[ A \right]^2}{\left[ B \right]^2}$ : This is incorrect. The reaction is zero order with respect to A, and the order with respect to B is 3.

Key Concepts:
- Rate Law
- Order of Reaction

Memory Aids:
- The rate law is determined experimentally.

Quick Tips:
- Compare experiments where only one concentration changes at a time.

25 Medium

25 of 71Question permalink

Understanding +4 | -1

Which of the following quantities changes on addition of a catalyst during a chemical reaction ?

View answer and explanation

Correct option: B

Activation energy

Option B (Activation energy) is correct because a catalyst lowers the activation energy of a reaction, thus speeding it up.

Solution approach

Correct Answer + Core Reason: Activation energy. A catalyst changes the activation energy of a reaction.

Step-by-Step Solution:

Step 1: Define a catalyst: A substance that speeds up a chemical reaction without being consumed in the process.

Step 2: Understand the role of activation energy: Activation energy is the minimum energy required for a reaction to occur.

Step 3: A catalyst provides an alternative reaction pathway with a lower activation energy.

Step 4: This lower activation energy allows more molecules to have sufficient energy to react, thus increasing the reaction rate.

Why other options are wrong:
- Option A: Equilibrium constant: A catalyst does not change the equilibrium constant of a reaction; it only affects the rate at which equilibrium is reached.
- Option C: Gibbs energy: A catalyst does not change the Gibbs free energy change of a reaction; it only affects the rate.
- Option D: Enthalpy: A catalyst does not change the enthalpy change of a reaction; it only affects the rate.

Key Concepts:
- Catalyst
- Activation energy
- Gibbs free energy
- Enthalpy
- Equilibrium constant

Memory Aids:
- Catalysts lower activation energy.

Quick Tips:
- Catalysts speed up reactions by lowering the activation energy but do not affect the thermodynamics of the reaction.

26 Medium

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Understanding +4 | -1

Which plots will give the value of activation energy?

View answer and explanation

Correct option: D

\ln \;K\;\mathrm{ }vs.\;\frac{1}{T}

Option D (

\ln \;K\;\mathrm{ }\;vs.\;\frac{1}{T}

) is correct because the slope of this plot is related to the activation energy according to the Arrhenius equation.

Solution approach

Correct Answer + Core Reason: $\ln \;K\;\mathrm{ }\;vs.\;\frac{1}{T}$ . The Arrhenius equation relates the rate constant to temperature and activation energy; a plot of ln K vs 1/T yields a straight line whose slope is related to the activation energy.

Step-by-Step Solution:

Step 1: Recall the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$ , where k is the rate constant, A is the pre-exponential factor, $E_a$ is the activation energy, R is the gas constant, and T is the temperature.

Step 2: Take the natural logarithm of both sides of the Arrhenius equation: $\ln k = \ln A - \frac{E_a}{RT}$

Step 3: Rearrange the equation to resemble the equation of a straight line (y = mx + c): $\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$

Step 4: Identify the variables: y = $\ln k$ , x = $\frac{1}{T}$ , m (slope) = $-\frac{E_a}{R}$ , and c (y-intercept) = $\ln A$

Step 5: The plot of $\ln K$ vs $\frac{1}{T}$ will give a straight line with a slope of $-\frac{E_a}{R}$ . Therefore, the activation energy can be determined from the slope of this plot.

Why other options are wrong:
- Option A: $K\mathrm{ }\;vs.\;T$ : This plot will not give a linear relationship and cannot be directly used to determine the activation energy.
- Option B: $1/K\;\mathrm{ }\;vs\;.T$ : This plot will also not give a linear relationship and cannot be directly used to determine the activation energy.
- Option C: $\ln \;K\;\mathrm{ }\;vs.\;T$ : This plot will not give a linear relationship. The Arrhenius equation indicates that $\ln K$ is linearly related to $1/T$ , not T.

Key Concepts:
- Arrhenius Equation
- Activation Energy
- Rate Constant
- Linear Plots

Memory Aids:
- Remember the Arrhenius equation and how to linearize it using logarithms.

Quick Tips:
- Always remember to take the natural logarithm of the rate constant when dealing with the Arrhenius equation.
- The slope of the $\ln K$ vs $1/T$ plot is negative, so the activation energy is positive.

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Application +4 | -1

{\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right) + 22\;\mathrm{kcal}

.

The activation energy for the forward reaction is

50\mathrm{ }\;kcal.

What is the activation energy for the backward reaction?

View answer and explanation

Correct option: D

+ 72\mathrm{ }\;kcal

Option D (

+ 72\mathrm{ }\;kcal

) is correct because the activation energy for the backward reaction is the sum of the activation energy for the forward reaction and the enthalpy change of the reaction.

Solution approach

Correct Answer + Core Reason: $+ 72\mathrm{ }\;kcal$ . The activation energy for the backward reaction is the sum of the forward activation energy and the enthalpy change.

Step-by-Step Solution:

Step 1: Identify the given information:\nActivation energy for the forward reaction, ${E_{a(f)}} = 50\mathrm{ }\;kcal$ \nEnthalpy change for the reaction, $\Delta H = -22\mathrm{ }\;kcal$ (exothermic reaction)

Step 2: Recall the relationship between activation energies and enthalpy change:
$\Delta H = {E_{a(f)}} - {E_{a(b)}}$ , where ${E_{a(b)}}$ is the activation energy for the backward reaction.

Step 3: Rearrange the equation to solve for ${E_{a(b)}}$ :
${E_{a(b)}} = {E_{a(f)}} - \Delta H$

Step 4: Substitute the given values:
${E_{a(b)}} = 50\mathrm{ }\;kcal - (-22\mathrm{ }\;kcal)$

Step 5: Calculate the activation energy for the backward reaction:
${E_{a(b)}} = 50\mathrm{ }\;kcal + 22\mathrm{ }\;kcal = 72\mathrm{ }\;kcal$

Final Answer: The activation energy for the backward reaction is $72\mathrm{ }\;kcal$ .

Why other options are wrong:
- Option A: $- 72\mathrm{ }\;kcal$ : This is incorrect because it subtracts the enthalpy change from the forward activation energy without considering the sign convention.
- Option B: $- 28\mathrm{ }\;kcal$ : This is incorrect because it subtracts the forward activation energy from the enthalpy change.
- Option C: $+ 28\mathrm{ }\;kcal$ : This is incorrect because it calculates the difference between the forward activation energy and the absolute value of the enthalpy change, but the correct formula requires considering the sign of $\Delta H$ .

Key Concepts:
- Activation energy
- Enthalpy change
- Relationship between forward and backward activation energies and enthalpy change

Memory Aids:
- Remember that for exothermic reactions, $\Delta H$ is negative, and for endothermic reactions, $\Delta H$ is positive.
- Visualize the energy diagram to understand the relationship between ${E_{a(f)}}$ \, ${E_{a(b)}}$ \, and $\Delta H$ .

Quick Tips:
- Always pay attention to the sign of $\Delta H$ when calculating activation energies.
- Draw an energy diagram to visualize the relationship between the activation energies and the enthalpy change.

28 Medium

28 of 71Question permalink

Analyzing +4 | -1

For a reaction between A and B, the initial rate of reaction is measured for various initial concentrations A and B. the data provided are

\begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\left[ A \right]\,\,\,\,\,\,\,\,\,\,\,\left[ B \right]\;\;\;\;\;\;\;Initial\mathrm{ }reaction\mathrm{ }rate\\ 1\,\,\,\,\,\,\,0.20M\,\,\,\,\,0.30M\,\,\,\,\,\,5 \times {10^{ - 5}}\\ 2\,\,\,\,\,0.20\mathrm{ }M\,\,\,\,0.10\mathrm{ }M\,\,\,\,\,\,5 \times {10^{ - 5}}\\ 3\,\,\,\,\,0.40\mathrm{ }M\,\,\,\,0.05\mathrm{ }M\,\,\,\,\,\,1 \times {10^{ - 5}} \end{array}

The overall order of the reaction is

View answer and explanation

Correct option: A

One

Option A (One) is correct because the rate is independent of [B] and proportional to [A], making the overall order 1.

Solution approach

Correct Answer + Core Reason: One. The rate is independent of [B] and proportional to [A], making the overall order 1.

Step-by-Step Solution:

Step 1: Write the general rate law: Rate = $k[A]^m[B]^n$ , where m and n are the orders with respect to A and B, respectively.

Step 2: Compare experiments 1 and 2: [A] is constant, and [B] changes from 0.30 M to 0.10 M. The initial reaction rate remains constant at $5 \times 10^{-5}$ . This indicates that the rate is independent of [B], so $n = 0$ .

Step 3: Compare experiments 2 and 3: [A] changes from 0.20 M to 0.40 M, and [B] changes from 0.10 M to 0.05 M. The initial reaction rate changes from $5 \times 10^{-5}$ to $1 \times 10^{-4}$ . Since we know that the rate is independent of [B], we can focus on the change in [A].

Step 4: Since the rate doubles when [A] doubles, the reaction is first order with respect to A. Therefore, $m = 1$ .

Step 5: The overall order of the reaction is $m + n = 1 + 0 = 1$ .

Why other options are wrong:
- Option B: This is incorrect because the reaction is not second order overall. The rate does not change proportionally to the square of the concentrations.
- Option C: This is incorrect because the reaction is not two and a half order overall. The rate does not change proportionally to the concentration raised to the power of 2.5.
- Option D: This is incorrect because the reaction is not third order overall. The rate does not change proportionally to the cube of the concentrations.

Key Concepts:
- Rate law
- Order of reaction
- Initial rate method

Memory Aids:
- The order of a reaction with respect to a particular reactant is the exponent of that reactant's concentration in the rate law.
- The overall order of the reaction is the sum of the orders with respect to each reactant.

Quick Tips:
- Use the initial rate method to determine the order of the reaction with respect to each reactant.
- Compare experiments where only one reactant concentration changes at a time.

29 Medium

29 of 71Question permalink

Application +4 | -1

The rate of the reaction

A \to products

, at the initial concentration of

3.24 \times {10^{ - 2}}\;M

is nine times its rate at another initial concentration of

1.2 \times {10^{ - 3}}\;M

. The order of the reaction is

View answer and explanation

Correct option: D

\frac{2}{3}

Option D (2/3) is correct because the order of the reaction is calculated using the rate law and given data.

Solution approach

Correct Answer + Core Reason: $\frac{2}{3}$ . The order of the reaction is calculated using rate law.

Step-by-Step Solution:

Step 1: Define the rate law: Assume the rate law is $rate = k[A]^n$ , where $n$ is the order of the reaction.

Step 2: Set up the rate equations using the given data:
$rate_1 = k(3.24 \times 10^{-2})^n$
$rate_2 = k(1.2 \times 10^{-3})^n$ \nGiven that $rate_1 = 9 \times rate_2$ , we have:
$k(3.24 \times 10^{-2})^n = 9 \times k(1.2 \times 10^{-3})^n$

Step 3: Simplify the equation:
$(\frac{3.24 \times 10^{-2}}{1.2 \times 10^{-3}})^n = 9$
$(27)^n = 9$

Step 4: Solve for n:\nSince $27 = 3^3$ and $9 = 3^2$ , we can rewrite the equation as:
$(3^3)^n = 3^2$
$3^{3n} = 3^2$ \nTherefore, $3n = 2$ , and $n = \frac{2}{3}$

Step 5: Final Answer: The order of the reaction is $\frac{2}{3}$ .

Why other options are wrong:
- Option A (1/2): This is incorrect because it does not satisfy the rate relationship given in the problem.
- Option B (3/4): This is incorrect because it does not satisfy the rate relationship given in the problem.
- Option C (3/2): This is incorrect because it does not satisfy the rate relationship given in the problem.

Key Concepts:
- Rate law
- Order of reaction

Memory Aids:
- The rate law expresses the relationship between the rate of a reaction and the concentrations of the reactants.

Quick Tips:
- When solving for the order of a reaction, set up rate equations using the given data and solve for the exponent.

30 Medium

30 of 71Question permalink

Remembering +4 | -1

The rate of reaction depends upon the

View answer and explanation

Correct option: D

Conc. of reactants

Option D is correct because the rate of a reaction is directly proportional to the concentration of the reactants, as described by the rate law.

Solution approach

Correct Answer + Core Reason: Conc. of reactants. The rate of a reaction depends on the concentration of the reactants.

Step-by-Step Solution:

Step 1: Define rate of reaction. The rate of a reaction is the change in concentration of reactants or products per unit time.

Step 2: Relate rate to concentration. The rate law expresses the relationship between the rate of a reaction and the concentrations of the reactants. In general, increasing the concentration of reactants increases the rate of the reaction.

Why other options are wrong:
- Option A: Volume can affect concentration, but it's the concentration itself that directly affects the rate.
- Option B: Force is not a direct factor in determining the rate of a reaction.
- Option C: Pressure can affect the concentration of gaseous reactants, but it's the concentration that directly affects the rate.

Key Concepts:
- Rate of reaction
- Rate law
- Concentration

Memory Aids:
- Rate depends on concentration.

Quick Tips:
- The rate law is experimentally determined and relates rate to reactant concentrations.

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