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 $NaOH$

CHEMICAL KINETICS MCQs for NEET UG CHEMISTRY — Mixed difficulty

1 Easy

1 of 198Question permalink

Understanding +4 | -1

If the reaction rate at a given temperature becomes slower then

View answer and explanation

Correct option: A

The free energy of activation is higher

Option A (The free energy of activation is higher) is correct because a slower reaction rate implies a higher activation energy barrier.

Solution approach

Correct Answer + Core Reason: The free energy of activation is higher. A higher activation energy leads to a slower reaction rate.

Step-by-Step Solution:

Step 1: Understand the relationship between reaction rate and activation energy.

Step 2: Recall that the rate of a reaction is inversely proportional to the activation energy. A higher activation energy means fewer molecules have enough energy to overcome the barrier and react.

Step 3: Relate activation energy to free energy of activation. The free energy of activation is the energy required for the reactants to reach the transition state.

Step 4: Conclude that if the reaction rate is slower, the free energy of activation must be higher.

Why other options are wrong:
- Option B: A lower free energy of activation would lead to a faster reaction rate, not slower.
- Option C: While entropy changes can affect reaction rates, it's not the direct cause of a slower rate at a given temperature. The primary factor is the activation energy.
- Option D: The initial concentration of reactants affects the rate, but a slower rate at a given temperature specifically points to a higher activation energy, not necessarily a change in initial concentration.

Key Concepts:
- Activation Energy
- Free Energy of Activation
- Reaction Rate
- Arrhenius Equation

Memory Aids:
- Higher activation energy = Slower reaction rate

Quick Tips:
- Remember that activation energy is the energy barrier that must be overcome for a reaction to occur.

2 Easy

2 of 198Question permalink

Understanding +4 | -1

\frac{1}{{{{\left[ A \right]}^2}}}

vs times are a straight line. Order of reaction is

View answer and explanation

Correct option: D

Third

The correct answer is Third because the integrated rate law for a third-order reaction has the form

\frac{1}{[A]^2} = kt + constant

, which gives a straight line when

\frac{1}{[A]^2}

is plotted against time.

Solution approach

Correct Answer + Core Reason: Third. A plot of $\frac{1}{{{{\left[ A \right]}^2}}}$ vs time gives a straight line for a third-order reaction.

Step-by-Step Solution:

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

Step 2: For a third-order reaction (n=3): Rate = $k[A]^3 = -\frac{d[A]}{dt}$

Step 3: Integrate the rate law: $\int \frac{d[A]}{[A]^3} = -\int k dt$
$\frac{-1}{2[A]^2} = -kt + C$ , where C is the constant of integration.
$\frac{1}{[A]^2} = 2kt + C'$

Step 4: This equation has the form y = mx + b, where y = $\frac{1}{[A]^2}$ , x = t, m = 2k, and b = C'. Therefore, a plot of $\frac{1}{[A]^2}$ vs time gives a straight line.

Why other options are wrong:
- Option A: For a first-order reaction, ln[A] vs time is a straight line.
- Option B: For a second-order reaction, $\frac{1}{[A]}$ vs time is a straight line.
- Option C: For a zero-order reaction, [A] vs time is a straight line.

Key Concepts:
- Integrated rate laws
- Order of reaction
- Third-order reaction

Memory Aids:
- Third-order: $\frac{1}{[A]^2}$ vs time is linear.

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

3 Medium

3 of 198Question 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.

4 Easy

4 of 198Question permalink

Understanding +4 | -1

Acid hydrolysis of sucrose is a

View answer and explanation

Correct option: A

Pseudo first order reaction

The correct answer is A (Pseudo first order reaction) because the acid hydrolysis of sucrose involves water in large excess, making it a pseudo-first-order reaction.

Solution approach

Correct Answer + Core Reason: Pseudo first order reaction. Acid hydrolysis of sucrose is a pseudo-first-order reaction because water is present in large excess, and its concentration remains nearly constant during the reaction.

Step-by-Step Solution:

Step 1: Write the reaction: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 + C_6H_{12}O_6$ (Sucrose + Water -> Glucose + Fructose)

Step 2: The rate law is Rate = $k[C_{12}H_{22}O_{11}][H_2O]$ . However, since water is in large excess, its concentration remains nearly constant.

Step 3: Therefore, the rate law can be approximated as Rate = $k'[C_{12}H_{22}O_{11}]$ , where $k' = k[H_2O]$ . This is a first-order rate law, making the reaction pseudo-first-order.

Why other options are wrong:
- Option B: It is not a zero-order reaction because the rate depends on the concentration of sucrose.
- Option C: It is not a second-order reaction because the concentration of water does not significantly change during the reaction.
- Option D: While it involves a single molecule of sucrose reacting, the presence of water in large excess makes it a pseudo-first-order reaction rather than a simple unimolecular reaction.

Key Concepts:
- Pseudo-first-order reaction
- Hydrolysis
- Rate law

Memory Aids:
- Pseudo-first-order reactions involve a reactant in large excess.

Quick Tips:
- Recognize that reactions with a reactant in large excess often follow pseudo-first-order kinetics.

5 Easy

5 of 198Question permalink

Remembering +4 | -1

Order of radioactive disintegration reaction is

View answer and explanation

Correct option: B

First

The correct answer is B (First) because radioactive decay follows first-order kinetics.

Solution approach

Correct Answer + Core Reason: First. Radioactive disintegration follows first-order kinetics.

Step-by-Step Solution:

Step 1: Radioactive decay is a spontaneous process.

Step 2: The rate of decay is proportional to the number of radioactive nuclei present: Rate = $k \times N$ , where N is the number of radioactive nuclei and k is the decay constant.

Step 3: This rate law corresponds to first-order kinetics.

Why other options are wrong:
- Option A: Radioactive decay does not follow zero-order kinetics.
- Option C: Radioactive decay does not follow second-order kinetics.
- Option D: Radioactive decay does not follow third-order kinetics.

Key Concepts:
- Radioactive decay
- First-order kinetics
- Decay constant

Memory Aids:
- Radioactive decay is always first order.

Quick Tips:
- Remember that radioactive decay follows first-order kinetics.

6 Medium

6 of 198Question 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 $x^n$ .

7 Easy

7 of 198Question permalink

Understanding +4 | -1

On addition of

\mathrm{AgN}{\mathrm{O}_3}\;to\;\mathrm{NaCl},

white ppt. occurs:

View answer and explanation

Correct option: A

Instantaneously

Option A (Instantaneously) is correct because the reaction between Ag+ and Cl- ions is a very fast ionic reaction.

Solution approach

Correct Answer + Core Reason: Instantaneously. The reaction between silver ions and chloride ions is a rapid ionic reaction.

Step-by-Step Solution:

Step 1: Identify the reactants and products. We have silver nitrate ( $\mathrm{AgNO}_3$ ) and sodium chloride ( $\mathrm{NaCl}$ ). The reaction produces silver chloride ( $\mathrm{AgCl}$ ) and sodium nitrate ( $\mathrm{NaNO}_3$ ).

Step 2: Write the balanced chemical equation: $\mathrm{AgNO}_3(aq) + \mathrm{NaCl}(aq) \rightarrow \mathrm{AgCl}(s) + \mathrm{NaNO}_3(aq)$

Step 3: Recognize that this is an ionic reaction. Silver ions ( $\mathrm{Ag}^+$ ) react with chloride ions ( $\mathrm{Cl}^-$ ) to form a precipitate of silver chloride ( $\mathrm{AgCl}$ ).

Step 4: Ionic reactions are generally very fast because they involve the combination of oppositely charged ions in solution.

Step 5: Therefore, the formation of the white precipitate of silver chloride occurs almost instantaneously.

Why other options are wrong:
- Option B: With a measurable speed: While all reactions have a speed, this reaction is so fast that measuring it without specialized equipment is difficult.
- Option C: Slowly: This is incorrect because ionic reactions are generally very fast.
- Option D: None of these: This is incorrect because the reaction does occur, and it occurs instantaneously.

Key Concepts:
- Ionic reactions
- Precipitation reactions
- Reaction rates

Memory Aids:
- "All nitrates are soluble" (except for a few rare exceptions). Silver chloride is a common insoluble salt.

Quick Tips:
- Remember that reactions between ions in solution are generally very fast.

8 Medium

8 of 198Question 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.

9 Hard

9 of 198Question permalink

Understanding +4 | -1

In the reaction

\mathrm{A} \longrightarrow \mathrm{P}

, concentration of

\mathrm{A}

was registered after equal intervals of time. If these values followed A.P., then order of reaction will be:

View answer and explanation

Correct option: A

0

Option A (0) is correct because if the concentration of A follows an arithmetic progression (A.P.) with equal time intervals, it indicates a zero-order reaction.

Solution approach

Correct Answer + Core Reason: 0. If the concentration of A follows an A.P., the reaction is zero order.

Step-by-Step Solution:

Step 1: Recall the integrated rate law for a zero-order reaction: $\left[ A \right] = {\left[ A \right]_0} - kt$

Step 2: This equation shows that the concentration of A decreases linearly with time.

Step 3: A linear decrease in concentration with equal time intervals implies that the concentrations form an arithmetic progression (A.P.).

Step 4: Therefore, if the concentrations follow an A.P., the reaction is zero order.

Why other options are wrong:
- Option B: For a first-order reaction, the logarithm of the concentration decreases linearly with time.
- Option C: For a second-order reaction, the inverse of the concentration increases linearly with time.
- Option D: Not applicable since all options are the same.

Key Concepts:
- Zero-order reaction
- Integrated rate law
- Arithmetic progression

Memory Aids:
- Linear decrease in concentration = Zero-order reaction

Quick Tips:
- Relate the integrated rate laws to the behavior of concentrations with time.

10 Easy

10 of 198Question permalink

Analyzing +4 | -1

The concentration of R in the reaction

R \to P

was measured as a function of time and the following data is obtained

\begin{array}{llllllllllllllllllll} {\left[ R \right]\left( {molar} \right)\;\;\;1.00\;\;\;\;\;\;0.75\;\;\;\;\;0.40\;\;\;\;\;0.10}\\ {t\;\left( {min} \right)\;\;\;\;\;\;\;\;\;0.00\;\;\;\;\;0.05\;\;\;\;\;0.12\;\;\;\;\;0.18} \end{array}\;

The order of the reaction is

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because the concentration decreases linearly with time, indicating a zero-order reaction.

Solution approach

Correct Answer + Core Reason: Zero. The reaction is zero order because the rate of disappearance of R is constant.

Step-by-Step Solution:

Step 1: Analyze the data to determine the order of the reaction.

Step 2: Check if it's zero order. For a zero-order reaction, the rate is constant, i.e., the concentration decreases linearly with time.

Step 3: Calculate the change in concentration over time intervals:
- From t = 0.00 to t = 0.05: $\Delta[R] = 1.00 - 0.75 = 0.25$ , $\Delta t = 0.05 - 0.00 = 0.05$ , Rate = $\frac{0.25}{0.05} = 5$
- From t = 0.05 to t = 0.12: $\Delta[R] = 0.75 - 0.40 = 0.35$ , $\Delta t = 0.12 - 0.05 = 0.07$ , Rate = $\frac{0.35}{0.07} = 5$
- From t = 0.12 to t = 0.18: $\Delta[R] = 0.40 - 0.10 = 0.30$ , $\Delta t = 0.18 - 0.12 = 0.06$ , Rate = $\frac{0.30}{0.06} = 5$

Step 4: Since the rate is constant, the reaction is zero order.

Why other options are wrong:
- Option B: First: For a first-order reaction, the half-life is constant, which is not evident from the data.
- Option C: Second: For a second-order reaction, the inverse of the concentration would change linearly with time, which is not the case here.
- Option D: Third: Third-order reactions have more complex relationships that are not supported by the data.

Key Concepts:
- Order of reaction
- Zero-order kinetics

Memory Aids:
- Zero-order: Rate is constant. First-order: Half-life is constant.

Quick Tips:
- Check for constant rate to identify zero-order reactions.

11 Hard

11 of 198Question permalink

Analyzing +4 | -1

For a homogeneous gaseous reaction

\mathrm{A} \longrightarrow 3 \mathrm{B}

, if pressure after time

\mathrm{t}

was

\mathrm{P}_{\mathrm{t}}

and after completion of reaction, pressure was

\mathrm{P}_{\infty}

then select correct relation

View answer and explanation

Correct option: D

\mathrm{k}=\frac{1}{\mathrm{t}} \ln \left(\frac{2 \mathrm{P}_{\infty}}{3\left(\mathrm{P}_{\infty}-\mathrm{P}_\mathrm{t}\right)}\right)

The correct answer is D (

\mathrm{k}=\frac{1}{\mathrm{t}} \ln \left(\frac{2 \mathrm{P}_{\infty}}{3\left(\mathrm{P}_{\infty}-\mathrm{P}_\mathrm{t}\right)}\right)

) because it correctly relates the rate constant to the pressures at time t and at completion for a first-order reaction.

Solution approach

Correct Answer + Core Reason: $\mathrm{k}=\frac{1}{\mathrm{t}} \ln \left(\frac{2 \mathrm{P}_{\infty}}{3\left(\mathrm{P}_{\infty}-\mathrm{P}_\mathrm{t}\right)}\right)$ . This equation correctly relates the rate constant to the partial pressures of the reactant and product at time t and at completion for a first-order reaction.

Step-by-Step Solution:

Step 1: Write the reaction: $\mathrm{A} \longrightarrow 3 \mathrm{B}$

Step 2: Initial pressure: $P_0 = P_A + P_B = P_0 + 0 = P_0$

Step 3: Pressure at time t: $P_t = P_A + P_B = (P_0 - p) + 3p = P_0 + 2p$ , where p is the decrease in pressure of A.

Step 4: Pressure at completion: $P_{\infty} = 0 + 3P_0 = 3P_0$ \nSo, $P_0 = \frac{P_{\infty}}{3}$

Step 5: From Step 3, $2p = P_t - P_0 = P_t - \frac{P_{\infty}}{3}$
$p = \frac{1}{2} (P_t - \frac{P_{\infty}}{3})$

Step 6: The pressure of A at time t is $P_A = P_0 - p = \frac{P_{\infty}}{3} - \frac{1}{2} (P_t - \frac{P_{\infty}}{3}) = \frac{P_{\infty}}{3} - \frac{P_t}{2} + \frac{P_{\infty}}{6} = \frac{P_{\infty}}{2} - \frac{P_t}{2} = \frac{P_{\infty} - P_t}{2}$

Step 7: For a first-order reaction, $k = \frac{1}{t} ln(\frac{P_0}{P_A}) = \frac{1}{t} ln(\frac{P_{\infty}/3}{(P_{\infty} - P_t)/2}) = \frac{1}{t} ln(\frac{2P_{\infty}}{3(P_{\infty} - P_t)})$

Why other options are wrong:
- Option A: Incorrect relationship between k and the pressures.
- Option B: Incorrect relationship between k and the pressures.
- Option C: Incorrect relationship between k and the pressures.

Key Concepts:
- First-order kinetics
- Partial pressures
- Rate constant

Memory Aids:
- $k = \frac{1}{t} ln(\frac{P_0}{P_A})$ for first-order gas-phase reactions

Quick Tips:
- Relate the partial pressures to the initial pressure and the extent of the reaction.
- Use the ideal gas law to relate pressure to concentration.

12 Medium

12 of 198Question 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.

13 Medium

13 of 198Question 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.

14 Hard

14 of 198Question permalink

Understanding +4 | -1

Which of the following statement(s) is/are correct?

(i) Rate of reaction decreases with passage of time as the concentration of reactants decrease.

(ii) For a reaction

\begin{aligned} & {pP}+{qQ} \longrightarrow {rR}+{sS} \\ & \text { Rate }={k}[{P}]^{{x}}[{Q}]^{{y}} \text { where } {x}={p} \text { and } {y}={q} \end{aligned}

(iii) Rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stichiometric coefficient of the reacting species in a balanced chemical equation.

View answer and explanation

Correct option: A

(i) and (iii)

Option A ((i) and (iii)) is correct because the rate of reaction decreases with time and the rate law is determined experimentally, not directly from stoichiometry.

Solution approach

Correct Answer + Core Reason: (i) and (iii). Rate decreases with time, and rate law is experimental.

Step-by-Step Solution:

Step 1: Analyze each statement:
- (i) Rate of reaction decreases with passage of time as the concentration of reactants decrease: This is correct. As reactants are consumed, the rate of reaction generally decreases.
- (ii) For a reaction ${pP}+{qQ} \longrightarrow {rR}+{sS}$ , Rate $={k}[{P}]^{{x}}[{Q}]^{{y}}$ where ${x}={p}$ and ${y}={q}$ : This is incorrect. The exponents $x$ and $y$ are determined experimentally and are not necessarily equal to the stoichiometric coefficients $p$ and $q$ .
- (iii) Rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stichiometric coefficient of the reacting species in a balanced chemical equation: This is correct. The rate law is determined experimentally.

Step 2: Based on the analysis, statements (i) and (iii) are correct.

Why other options are wrong:
- Option B: (i) and (ii): Statement (ii) is incorrect.
- Option C: (ii) and (iii): Statement (ii) is incorrect.
- Option D: (i) only: Statement (iii) is also correct.

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

Memory Aids:
- Remember that the rate law is determined experimentally.

Quick Tips:
- Understand the difference between the rate law and the stoichiometric coefficients.

15 Medium

15 of 198Question 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.

16 Medium

16 of 198Question 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.

17 Medium

17 of 198Question 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.

18 Medium

18 of 198Question 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.

19 Medium

19 of 198Question 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.

20 Easy

20 of 198Question permalink

Application +4 | -1

In the following reaction

A \to B + C,

rate constant is

0.001\;\mathrm{M}{\mathrm{s}^{ - 1}}\mathrm{.}

Half-life and completion time of the given reaction are

View answer and explanation

Correct option: A

500 s, 1000 s

Option A (500 s, 1000 s) is correct because for a zero-order reaction, the half-life is a/2k and the completion time is a/k, where a is the initial concentration and k is the rate constant. Assuming initial concentration is 1M, half-life is 500s and completion time is 1000s.

Solution approach

Correct Answer + Core Reason: 500 s, 1000 s. For a zero-order reaction, half-life is $\frac{a}{2k}$ and completion time is $\frac{a}{k}$ .

Step-by-Step Solution:

Step 1: Identify the reaction order. Since the rate constant is given in M/s, the reaction is zero order.

Step 2: For a zero-order reaction, the rate law is: Rate = $k$ .

Step 3: The half-life ( $t_{1/2}$ ) of a zero-order reaction is given by: $t_{1/2} = \frac{[A]_0}{2k}$ , where $[A]_0$ is the initial concentration.

Step 4: The completion time ( $t_{completion}$ ) of a zero-order reaction is given by: $t_{completion} = \frac{[A]_0}{k}$ .

Step 5: Assume the initial concentration $[A]_0 = 1$ M. Then, $t_{1/2} = \frac{1}{2 \times 0.001} = 500$ s, and $t_{completion} = \frac{1}{0.001} = 1000$ s.

Why other options are wrong:
- Option B: 500 s, 750 s: The completion time is incorrect.
- Option C: 250 s, 500 s: Both half-life and completion time are incorrect.
- Option D: 300 s, 600 s: Both half-life and completion time are incorrect.

Key Concepts:
- Zero-Order Kinetics
- Half-Life
- Completion Time

Memory Aids:
- Remember the formulas for half-life and completion time for zero-order reactions.

Quick Tips:
- Identify the reaction order from the units of the rate constant.

21 Medium

21 of 198Question 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.

22 Hard

22 of 198Question permalink

Application +4 | -1

The velocity constant of a reaction at

290\;K

was found to be

3.2 \times {10^{ - 3}}\;at\;300\;K

, it will be

View answer and explanation

Correct option: A

6.4 \times {10^{ - 3}}

Option A (

6.4 \times {10^{ - 3}}

) is correct because the rate constant increases with temperature, and it can be estimated using the Arrhenius equation or a simplified approach.

Solution approach

Correct Answer + Core Reason: $6.4 \times {10^{ - 3}}$ . The rate constant increases with temperature.

Step-by-Step Solution:

Step 1: Recognize that the temperature increased from 290 K to 300 K. Since the temperature increased, the rate constant should also increase.

Step 2: Since the temperature change is relatively small, we can approximate that the rate constant roughly doubles for every 10-degree Celsius increase. This is a rule of thumb, not a precise calculation.

Step 3: Given the rate constant at 290 K is $3.2 \times 10^{-3}$ , at 300 K it should be approximately double that value.

Step 4: Calculate: $2 \times (3.2 \times 10^{-3}) = 6.4 \times 10^{-3}$

Step 5: Final Answer: $6.4 \times 10^{-3}$

Why other options are wrong:
- **Option B ( $1.28 \times {10^{ - 2}}$ ): This value is too high. It suggests a much larger increase than expected for a 10 K increase in temperature.
- Option C ( $9.6 \times {10^{ - 3}}$ ): This value is also too high, indicating a more than doubling of the rate constant for a small temperature change.
- Option D ( $3.2 \times {10^{ - 4}}$ ): This value is much too low. The rate constant should increase, not decrease, with an increase in temperature.

Key Concepts:
- Arrhenius Equation
- Temperature dependence of reaction rates

Memory Aids:
- Rule of thumb: Rate constant roughly doubles for every 10°C increase in temperature.

Quick Tips:**
- For small temperature changes, estimate the change in rate constant rather than performing complex calculations.

23 Hard

23 of 198Question permalink

Analyzing +4 | -1

The rate of the reaction

2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2

can be written in three ways:

\begin{aligned} & \frac{-\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}={k}\left[\mathrm{~N}_2 \mathrm{O}_5\right] ; \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}={k}^{\prime}\left[\mathrm{N}_2 \mathrm{O}_5\right] \\ & \frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}={k}^{\prime \prime}\left[\mathrm{N}_2 \mathrm{O}_5\right] \end{aligned}

The relationship between

{k}

and

{k}^{\prime}

and between

{k}

and

{k}^{\prime \prime}

are:

View answer and explanation

Correct option: B

k^{\prime}=2 k ; k^{\prime \prime}=k / 2

The correct answer is B (

k^{\prime}=2 k ; k^{\prime \prime}=k / 2

) because the rate constants are related by the stoichiometry of the reaction.

Solution approach

Correct Answer + Core Reason: $k^{\prime}=2 k ; k^{\prime \prime}=k / 2$ . The rate constants are related to each other based on the stoichiometry of the balanced chemical equation.

Step-by-Step Solution:

Step 1: Write the rate of the reaction in terms of the change in concentration of each species:
$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$

Step 2: Substitute the given rate expressions:
$-\frac{1}{2} (k[N_2O_5]) = \frac{1}{4} (k'[N_2O_5]) = k''[N_2O_5]$

Step 3: Relate k' to k:
$-\frac{1}{2} k = \frac{1}{4} k'$
$k' = -2k \times -1 = 2k$

Step 4: Relate k'' to k:
$-\frac{1}{2} k = k''$
$k'' = \frac{k}{2}$

Step 5: Therefore, the relationships are:
$k' = 2k$ and $k'' = \frac{k}{2}$

Why other options are wrong:
- Option A: Incorrect relationship between k and k''.
- Option C: Incorrect relationship between k and k''.
- Option D: Incorrect relationship between k and k' and k and k''.

Key Concepts:
- Rate of reaction
- Rate constant
- Stoichiometry and reaction rates

Memory Aids:
- The rate of reaction is always positive. The negative sign is used for reactants because their concentration decreases with time.

Quick Tips:
- Always balance the chemical equation before relating the rate constants.
- Pay attention to the stoichiometric coefficients when relating the rates of different species.

24 Medium

24 of 198Question 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.

25 Easy

25 of 198Question permalink

Remembering +4 | -1

Temperature dependent equation can be written as

View answer and explanation

Correct option: D

None of these

Option D is correct because the temperature-dependent equation, derived from the Arrhenius equation, is

\ln k = \ln A - \frac{{E_a}}{{RT}}

.

Solution approach

Correct Answer + Core Reason: None of these. The correct temperature-dependent equation is derived from the Arrhenius equation and involves a negative exponential term.

Step-by-Step Solution:

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

Step 2: Take the natural logarithm of both sides: $\ln k = \ln (A e^{-E_a/RT})$

Step 3: Simplify using logarithm properties: $\ln k = \ln A + \ln (e^{-E_a/RT})$

Step 4: Further simplify: $\ln k = \ln A - \frac{E_a}{RT}$

Step 5: Compare with the given options. None of the options match the correct equation.

Why other options are wrong:
- Option A: $\mathrm{In}\;k = \mathrm{In}\;A - {e^{{E_a}/RT}}$ is incorrect because the exponential term should not be present after taking the logarithm.
- Option B: $\mathrm{In}\;k = \mathrm{In}\;A + {e^{{E_a}/RT}}$ is incorrect because the sign of the exponential term is wrong.
- Option C: $\mathrm{In}\;k = \mathrm{In}\;A - {e^{RT/{E_a}}}$ is incorrect because the exponential term is wrong and the ratio is inverted.

Key Concepts:
- Arrhenius equation
- Natural logarithm
- Temperature dependence of reaction rates

Memory Aids:
- Remember to take the natural logarithm of the Arrhenius equation.

Quick Tips:
- The logarithmic form of the Arrhenius equation is useful for plotting and determining activation energy.

26 Easy

26 of 198Question permalink

Remembering +4 | -1

Rate of reaction depends upon

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the rate of a reaction is influenced by temperature, catalysts, and concentration.

Solution approach

Correct Answer + Core Reason: All of these. The rate of a reaction depends on temperature, catalyst, and concentration.

Step-by-Step Solution:

Step 1: Understand the factors affecting reaction rates.

Step 2: Consider temperature. According to collision theory and the Arrhenius equation, increasing the temperature increases the kinetic energy of molecules, leading to more frequent and effective collisions, thus increasing the reaction rate.

Step 3: Consider catalyst. Catalysts provide an alternative reaction pathway with a lower activation energy, which increases the reaction rate without being consumed in the reaction.

Step 4: Consider concentration. According to the rate law, the rate of a reaction is often directly proportional to the concentration of the reactants. Increasing the concentration increases the frequency of collisions between reactant molecules, thus increasing the reaction rate.

Step 5: Conclusion. All three factors (temperature, catalyst, and concentration) affect the rate of a reaction.

Why other options are wrong:
- Option A (temperature): While temperature does affect the reaction rate, it's not the only factor.
- Option B (catalyst): While a catalyst does affect the reaction rate, it's not the only factor.
- Option C (concentration): While concentration does affect the reaction rate, it's not the only factor.

Key Concepts:
- Factors affecting reaction rates
- Temperature dependence (Arrhenius equation)
- Catalysis
- Concentration dependence (rate law)

Memory Aids:
- Remember that temperature, catalyst, and concentration are all important factors in determining the rate of a reaction.

Quick Tips:
- Consider all factors when analyzing reaction rates.

27 Hard

27 of 198Question permalink

Understanding +4 | -1

For the following

S_N

reaction,

\left(\mathrm{CH}_3\right)_3 \mathrm{CCl}+\mathrm{H}_2 \mathrm{O} \longrightarrow\left(\mathrm{CH}_3\right)_3 \mathrm{COH}+\mathrm{HCl}

Rate-determining step is

View answer and explanation

Correct option: A

\left(\mathrm{CH}_3\right)_3 \mathrm{CCl} \longrightarrow\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}+\mathrm{Cl}^{-}

Option A ((

\mathrm{CH}_3)_3 \mathrm{CCl} \longrightarrow(\mathrm{CH}_3)_3 \mathrm{C}^{+}+\mathrm{Cl}^{-}

) is correct because the rate-determining step in an SN1 reaction is the formation of the carbocation.

Solution approach

Correct Answer + Core Reason: $(\mathrm{CH}_3)_3 \mathrm{CCl} \longrightarrow(\mathrm{CH}_3)_3 \mathrm{C}^{+}+{\mathrm{Cl}^{-}}$ . The rate-determining step in an SN1 reaction is the formation of the carbocation.

Step-by-Step Solution:

Step 1: Recognize that this is an $S_N1$ reaction because the substrate is a tertiary alkyl halide.

Step 2: The $S_N1$ reaction proceeds in two steps:
- Step 1: Formation of a carbocation (rate-determining step).
- Step 2: Nucleophilic attack by water.

Step 3: The rate-determining step is the slow step, which is the formation of the carbocation: $(\mathrm{CH}_3)_3 \mathrm{CCl} \longrightarrow(\mathrm{CH}_3)_3 \mathrm{C}^{+}+{\mathrm{Cl}^{-}}$

Why other options are wrong:
- Option B: $(\mathrm{CH}_3)_3 \mathrm{CCl}+{\mathrm{H}_2}\mathrm{O} \longrightarrow(\mathrm{CH}_3)_3 \mathrm{COH}+\mathrm{HCl}$ : This is the overall reaction, not the rate-determining step.
- Option C: $(\mathrm{CH}_3)_3 \mathrm{C}^{+}+{\mathrm{H}_2}\mathrm{O} \longrightarrow(\mathrm{CH}_3)_3 \mathrm{COH}+{\mathrm{H}^{+}}$ : This is the second step of the $S_N1$ reaction, which is fast.
- Option D: $\mathrm{H}^{+}+{\mathrm{Cl}^-} \longrightarrow \mathrm{HCl}$ : This step is not part of the $S_N1$ mechanism.

Key Concepts:
- $S_N1$ reaction
- Rate-determining step
- Carbocation

Memory Aids:
- $S_N1$ : First step is rate-determining.

Quick Tips:
- Identify the type of reaction before determining the rate-determining step.

28 Easy

28 of 198Question permalink

Remembering +4 | -1

Half-life period of second order reaction is

View answer and explanation

Correct option: C

Inversely proportional to initial concentration of reactants

Option C (Inversely proportional to initial concentration of reactants) is correct because the half-life of a second-order reaction is inversely proportional to the initial concentration.

Solution approach

Correct Answer + Core Reason: Inversely proportional to initial concentration of reactants. The half-life of a second-order reaction is inversely proportional to the initial concentration.

Step-by-Step Solution:

Step 1: Recall the rate law for a second-order reaction: Rate = $k[A]^2$ , where k is the rate constant and [A] is the concentration of the reactant.

Step 2: The integrated rate law for a second-order reaction is: $\frac{1}{{[A]_t}} = \frac{1}{{[A]_0}} + kt$ , where $[A]_t$ is the concentration at time t, $[A]_0$ is the initial concentration, k is the rate constant, and t is time.

Step 3: Define half-life ( $t_{1/2}$ ) as the time it takes for the concentration to reduce to half of its initial value: $[A]_{t_{1/2}} = \frac{1}{2}[A]_0$ .

Step 4: Substitute this into the integrated rate law: $\frac{1}{{\frac{1}{2}[A]_0}} = \frac{1}{{[A]_0}} + kt_{1/2}$ .

Step 5: Simplify: $\frac{2}{{[A]_0}} = \frac{1}{{[A]_0}} + kt_{1/2}$ .

Step 6: Solve for $t_{1/2}$ : $t_{1/2} = \frac{1}{{k[A]_0}}$ .

Step 7: Conclusion: The half-life is inversely proportional to the initial concentration of the reactants.

Why other options are wrong:
- Option A (Proportional to initial concentration of reactants): This is incorrect for a second-order reaction.
- Option B (Independent of initial concentration of reactants): This is true for a first-order reaction, not a second-order reaction.
- Option D (None of the above): This is incorrect because option C is correct.

Key Concepts:
- Second-order kinetics
- Half-life of a reaction
- Integrated rate law

Memory Aids:
- Remember the relationship between half-life and initial concentration for different reaction orders.

Quick Tips:
- Memorize the formula for the half-life of a second-order reaction.

29 Easy

29 of 198Question permalink

Application +4 | -1

The activation energy of a reaction at a given temperature is found to be

2.303\;R\mathrm{T}\;\mathrm{ J}\;\mathrm{ mo}{\mathrm{l}^{ - 1}}

. The ratio of rate constant to the Arrhenius factor is

View answer and explanation

Correct option: B

0.1

Option B (0.1) is correct because the ratio of the rate constant to the Arrhenius factor is equal to

e^{-\frac{E_a}{RT}}

, and with the given activation energy, this simplifies to

e^{-2.303} = 0.1

.

Solution approach

Correct Answer + Core Reason: 0.1. The ratio of the rate constant to the Arrhenius factor is $e^{-\frac{E_a}{RT}}$ .

Step-by-Step Solution:

Step 1: Recall the Arrhenius equation: $k = Ae^{-\frac{E_a}{RT}}$ , where $k$ is the rate constant and $A$ is the Arrhenius factor.

Step 2: The ratio of the rate constant to the Arrhenius factor is $\frac{k}{A} = e^{-\frac{E_a}{RT}}$ .

Step 3: Substitute the given activation energy: $E_a = 2.303RT$

Step 4: Calculate the ratio: $\frac{k}{A} = e^{-\frac{2.303RT}{RT}} = e^{-2.303}$

Step 5: Since $e^{2.303} \approx 10$ , $e^{-2.303} = \frac{1}{10} = 0.1$ .

Why other options are wrong:
- Option A: $0.01$ : This would be the result if the activation energy was $4.606RT$ .
- Option C: $0.02$ : This is incorrect.
- Option D: $0.001$ : This is incorrect.

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

Memory Aids:
- Remember the Arrhenius equation: $k = Ae^{-\frac{E_a}{RT}}$

Quick Tips:
- Simplify the expression before calculating the final answer.

30 Medium

30 of 198Question 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.

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