For the reaction ${N_2}{O_5}(g) \to 2N{O_2}(g)\; + \;\frac{1}{2}{O_2}(g),$ the rate of disappearance of ${N_2}{O_5}$ is $6.25 \times {10^{ - 3}}\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}.$ The rate of formation of $N{O_2}$ and ${O_2}$ will be respectively.

$1.25 \times {10^{ - 2}}\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}\;and\;3.125 \times {10^{ - 3}}\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}$

$6.25 \times {10^{ - 3}}\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}\;and\;6.25 \times {10^{ - 3}}\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}$

$6.25 \times {10^{ - 3}}\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}\;and\;3.125 \times {10^{ - 3}}\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}$

$1.25 \times {10^{ - 2}}\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}\;and\;6.25 \times {10^{ - 3}}\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}$

Slope and intercepts of the plots ${\rm{ }}lo{g_{10}}\;K\;{\rm{vs}}\;\frac{1}{T}$ are given respectively by :

$ - \frac{{\Delta {H^ \circ }}}{R},\frac{{\Delta {S^ \circ }}}{R}$

$\frac{{\Delta {H^ \circ }}}{{2.303}},\frac{{ - \Delta {S^ \circ }}}{{2.303\;R}}$

CHEMICAL KINETICS MCQs for NEET UG CHEMISTRY — Easy

1 Easy

1 of 72Question 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 72Question 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 Easy

3 of 72Question 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.

4 Easy

4 of 72Question 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.

5 Easy

5 of 72Question 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.

6 Easy

6 of 72Question 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.

7 Easy

7 of 72Question 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.

8 Easy

8 of 72Question 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.

9 Easy

9 of 72Question 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.

10 Easy

10 of 72Question 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.

11 Easy

11 of 72Question 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.

12 Easy

12 of 72Question permalink

Remembering +4 | -1

In Arrhenius equation, the fraction of effective collisions is given by

View answer and explanation

Correct option: C

\mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}

Option C (

\mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}

) is correct because it represents the fraction of molecules with sufficient energy to overcome the activation energy, as defined by the Arrhenius equation.

Solution approach

Correct Answer + Core Reason: $\mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$ represents the fraction of effective collisions in the Arrhenius equation.

Step-by-Step Solution:

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

Step 2: Identify the term representing the fraction of effective collisions. This term is the exponential factor, $e^{-\frac{E_a}{RT}}$ . This factor indicates the proportion of molecules possessing energy equal to or greater than the activation energy, thus leading to successful collisions and a reaction.

Step 3: Conclude that the fraction of effective collisions is represented by $e^{-\frac{E_a}{RT}}$ .

Why other options are wrong:
- Option A: $\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$ is the entire Arrhenius equation, not just the fraction of effective collisions.
- Option B: $\mathrm{A}$ is the pre-exponential factor or frequency factor, representing the total number of collisions, not just the effective ones.
- Option D: $\mathrm{RT}$ is the product of the gas constant and temperature, which is related to energy but doesn't directly represent the fraction of effective collisions.

Key Concepts:
- Arrhenius Equation
- Activation Energy
- Effective Collisions

Memory Aids:
- Remember the Arrhenius equation: k = A exp(-Ea/RT)
- The exponential term relates to the fraction of molecules with sufficient energy.

Quick Tips:
- Understand the components of the Arrhenius equation and their physical meanings.
- Recognize that the exponential term is always a fraction between 0 and 1.

13 Easy

13 of 72Question permalink

Understanding +4 | -1

Effective collisions are those in which molecules must:

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because effective collisions require molecules to have sufficient energy (equal to or greater than the threshold energy), proper orientation, and acquire the activation energy to overcome the potential barrier.

Solution approach

Correct Answer + Core Reason: All of the above. Effective collisions require sufficient energy and proper orientation.

Step-by-Step Solution:

Step 1: Define effective collisions: Collisions that lead to a chemical reaction.

Step 2: List the requirements for effective collisions:
- Have energy equal to or greater than the threshold energy (A)
- Have proper orientation (B)
- Acquire the energy of activation (C)

Step 3: Combine the requirements: All the listed conditions must be met for a collision to be effective.

Why other options are wrong:
- Option A: Have energy equal to or greater than the threshold energy: This is a necessary condition, but not sufficient on its own.
- Option B: Have proper orientation: This is a necessary condition, but not sufficient on its own.
- Option C: Acquire the energy of activation: This is a necessary condition, but not sufficient on its own.

Key Concepts:
- Effective collisions
- Threshold energy
- Activation energy
- Proper orientation

Memory Aids:
- Remember the three key requirements for effective collisions: energy, orientation, and activation.

Quick Tips:
- Consider all factors that influence the success of a collision.

14 Easy

14 of 72Question permalink

Remembering +4 | -1

The inversion of cane sugar into glucose and fructose is:

View answer and explanation

Correct option: A

I order

The correct answer is A (I order) because the inversion of cane sugar is a pseudo-first-order reaction, where the concentration of water is in large excess.

Solution approach

Correct Answer + Core Reason: I order. The inversion of cane sugar is a pseudo-first-order reaction because the concentration of water is in large excess and remains nearly constant.

Step-by-Step Solution:

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

Step 2: The rate law is Rate = $k[C_{12}H_{22}O_{11}]^1[H_2O]^0$ because water is in large excess and its concentration remains nearly constant.

Step 3: Since the concentration of water is essentially constant, the reaction behaves as a first-order reaction with respect to sucrose.

Why other options are wrong:
- Option B: It is not a second-order reaction because the concentration of water does not significantly change during the reaction.
- Option C: It is not a third-order reaction.
- Option D: It is not a zero-order reaction because the rate depends on the concentration of sucrose.

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

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.

15 Easy

15 of 72Question permalink

Remembering +4 | -1

Which of the following theory is not related to chemical kinetics?

View answer and explanation

Correct option: D

VSPER theory

Option D (VSPER theory) is correct because VSEPR theory deals with molecular geometry, not reaction rates or mechanisms, which are the focus of chemical kinetics.

Solution approach

Correct Answer + Core Reason: VSPER theory is not related to chemical kinetics as it focuses on molecular geometry rather than reaction rates.

Step-by-Step Solution:

Step 1: Identify the theories related to chemical kinetics.

Step 2: Collision theory, activated complex theory, and absolute reaction rate theory are all used to explain reaction rates and mechanisms.

Step 3: VSEPR theory (Valence Shell Electron Pair Repulsion theory) is used to predict the shapes of molecules based on the repulsion between electron pairs around a central atom.

Step 4: Therefore, VSEPR theory is not related to chemical kinetics.

Why other options are wrong:
- Option A: Collision theory explains reaction rates based on the frequency and effectiveness of collisions between molecules.
- Option B: Activated complex theory describes the transition state of a reaction and its energy.
- Option C: Absolute reaction rate theory (also known as transition state theory) provides a theoretical framework for calculating reaction rates based on the properties of the activated complex.
- Option D: VSEPR theory is about molecular shapes, not reaction rates.

Key Concepts:
- Chemical kinetics
- Collision theory
- Activated complex theory
- Absolute reaction rate theory
- VSEPR theory

Memory Aids:
- VSEPR: Valence Shell Electron Pair Repulsion

Quick Tips:
- Remember that chemical kinetics deals with reaction rates and mechanisms, while VSEPR theory deals with molecular geometry.

16 Easy

16 of 72Question permalink

Analyzing +4 | -1

2{N_2}{O_5}\;\leftrightharpoons\;4N{O_2} + {O_2}

For the above reaction which of the following is not correct above rates of reaction?

View answer and explanation

Correct option: D

\frac{{ - 2d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 4\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}

Option D (

\frac{{ - 2d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 4\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}

) is correct because the rates are not correctly related according to the stoichiometry of the reaction.

Solution approach

Correct Answer + Core Reason: $\frac{{ - 2d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 4\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}$ . The rate expression should reflect the stoichiometry of the reaction.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2{N_2}{O_5} \rightleftharpoons 4N{O_2} + {O_2}$ .

Step 2: Define the rate of the reaction in terms of the change in concentration of reactants and products with time:\nRate = $- \frac{1}{2} \frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{1}{4} \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}$ .

Step 3: Analyze each option:
- Option A: $\frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 2\frac{{d\left[ {{O_2}} \right]}}{{dt}}$ => $- \frac{1}{2} \frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}$ . This is correct.
- Option B: $\frac{{ - 2d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{{d\left[ {N{O_2}} \right]}}{{dt}}$ => $- \frac{1}{2} \frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{1}{4} \frac{{d\left[ {N{O_2}} \right]}}{{dt}}$ . This is correct.
- Option C: $\frac{{d\left[ {N{O_5}} \right]}}{{dt}} = 4\frac{{d\left[ {{O_2}} \right]}}{{dt}}$ => $- \frac{1}{2} \frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}$ . Incorrect. It should be $- \frac{1}{2} \frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}$ .
- Option D: $\frac{{ - 2d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 4\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}$ is incorrect because it does not follow the correct stoichiometric relationship.

Why other options are wrong:
- Option A: $\frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 2\frac{{d\left[ {{O_2}} \right]}}{{dt}}$ : This is a correct representation of the rate.
- Option B: $\frac{{ - 2d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{{d\left[ {N{O_2}} \right]}}{{dt}}$ : This is a correct representation of the rate.
- Option C: $\frac{{d\left[ {N{O_5}} \right]}}{{dt}} = 4\frac{{d\left[ {{O_2}} \right]}}{{dt}}$ : This is a correct representation of the rate.

Key Concepts:
- Rate of Reaction
- Stoichiometry

Memory Aids:
- Remember to divide the rate of change of each species by its stoichiometric coefficient.

Quick Tips:
- Always balance the chemical equation before writing the rate expression.

17 Easy

17 of 72Question permalink

Application +4 | -1

For the reaction

{N_2}{O_5}(g) \to 2N{O_2}(g)\; + \;\frac{1}{2}{O_2}(g),

the rate of disappearance of

{N_2}{O_5}

is

6.25 \times {10^{ - 3}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}.

The rate of formation of

N{O_2}

and

{O_2}

will be respectively.

View answer and explanation

Correct option: B

1.25 \times {10^{ - 2}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}\;and\;3.125 \times {10^{ - 3}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}

Option B (

1.25 \times {10^{ - 2}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}\;and\;3.125 \times {10^{ - 3}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}

) is correct because it correctly calculates the rates of formation of

NO_2

and

O_2

based on the stoichiometry of the reaction.

Solution approach

Correct Answer + Core Reason: $1.25 \times {10^{ - 2}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}\;and\;3.125 \times {10^{ - 3}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}$ . The rates of formation are calculated based on the stoichiometry of the reaction and the rate of disappearance of $N_2O_5$ .

Step-by-Step Solution:

Step 1: Write the rate expression for the reaction:\nRate = $-\frac{d[N_2O_5]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt} = 2 \frac{d[O_2]}{dt}$

Step 2: We are given that $-\frac{d[N_2O_5]}{dt} = 6.25 \times 10^{-3} mol L^{-1} s^{-1}$

Step 3: We need to find $\frac{d[NO_2]}{dt}$ and $\frac{d[O_2]}{dt}$ .

Step 4: $\frac{d[NO_2]}{dt} = 2 \times (-\frac{d[N_2O_5]}{dt}) = 2 \times 6.25 \times 10^{-3} mol L^{-1} s^{-1} = 1.25 \times 10^{-2} mol L^{-1} s^{-1}$

Step 5: $\frac{d[O_2]}{dt} = \frac{1}{2} \times (-\frac{d[N_2O_5]}{dt}) = \frac{1}{2} \times 6.25 \times 10^{-3} mol L^{-1} s^{-1} = 3.125 \times 10^{-3} mol L^{-1} s^{-1}$

Step 6: Final Answer: $1.25 \times 10^{-2} mol L^{-1} s^{-1}$ and $3.125 \times 10^{-3} mol L^{-1} s^{-1}$

Why other options are wrong:
- Option A: This is incorrect because the rates of formation are not calculated correctly using the stoichiometric coefficients.
- Option C: This is incorrect because the rate of formation of $NO_2$ is not calculated correctly.
- Option D: This is incorrect because the rate of formation of $O_2$ is not calculated correctly.

Key Concepts:
- Rate of reaction
- Stoichiometry and rate expression

Memory Aids:
- Remember to multiply/divide the rate of formation/disappearance by the stoichiometric coefficient when relating the rates of different species in a reaction.

Quick Tips:
- Always write the balanced chemical equation and the corresponding rate expression before solving the problem.

18 Easy

18 of 72Question permalink

Analyzing +4 | -1

The following homogeneous gaseous reactions were experimentally found to be second order overall.

1.\;2\mathrm{NO} \to {\mathrm{N}_2} + {\mathrm{O}_2}

2.\;3{\mathrm{O}_2} \to 2{\mathrm{O}_3}

\mathrm{3}\mathrm{.}\;{\mathrm{N}_2}{\mathrm{O}_3} \to \mathrm{NO} + \mathrm{N}{\mathrm{O}_2}

\mathrm{4}\mathrm{.}\;{\mathrm{H}_2} + {\mathrm{I}_2} \to 2\mathrm{HI}

Which of these are most likely to be elementary reaction that occur in one step?

View answer and explanation

Correct option: C

1 and 4

Option C (1 and 4) is correct because bimolecular elementary reactions are more common and likely to occur in one step.

Solution approach

Correct Answer + Core Reason: 1 and 4. Elementary reactions that occur in one step are more likely to be bimolecular (involving two molecules).

Step-by-Step Solution:

Step 1: Analyze each reaction:
- 1: $2\mathrm{NO} \to {\mathrm{N}_2} + {\mathrm{O}_2}$ : This reaction involves two NO molecules, making it bimolecular and possibly elementary.
- 2: $3{\mathrm{O}_2} \to 2{\mathrm{O}_3}$ : This reaction involves three oxygen molecules, making it termolecular. Termolecular reactions are rare as they require simultaneous collision of three molecules.
- 3: ${\mathrm{N}_2}{\mathrm{O}_3} \to \mathrm{NO} + \mathrm{N}{\mathrm{O}_2}$ : This reaction involves a single molecule decomposing, making it unimolecular and possibly elementary.
- 4: ${\mathrm{H}_2} + {\mathrm{I}_2} \to 2\mathrm{HI}$ : This reaction involves two molecules (H2 and I2), making it bimolecular and possibly elementary.

Step 2: Given that the overall reaction is second order, the elementary reactions are more likely to be bimolecular.

Step 3: Reactions 1 and 4 are bimolecular and thus more likely to be elementary.

Why other options are wrong:
- Option A: 3 only: Reaction 3 is unimolecular, but we need to consider the second-order overall reaction.
- Option B: 1 and 3: Reaction 3 is unimolecular, making this option less likely.
- Option D: 3 and 4: Reaction 3 is unimolecular, making this option less likely.

Key Concepts:
- Elementary reaction
- Molecularity
- Order of reaction

Memory Aids:
- Unimolecular: one molecule, Bimolecular: two molecules, Termolecular: three molecules.

Quick Tips:
- Elementary reactions are usually unimolecular or bimolecular.
- Termolecular reactions are rare.

19 Easy

19 of 72Question permalink

Understanding +4 | -1

Slope and intercepts of the plots

\mathrm{ }lo{g_{10}}\;K\;\mathrm{vs}\;\frac{1}{T}

are given respectively by :

View answer and explanation

Correct option: C

\frac{{ - \Delta {H^ \circ }}}{{2.303\;R}},\frac{{\Delta {S^ \circ }}}{{2.303\;R}}

Option C is correct because the slope and intercept of the plot log₁₀K vs 1/T are given by -ΔH°/(2.303R) and ΔS°/(2.303R) respectively, according to the van't Hoff equation.

Solution approach

Correct Answer + Core Reason: $\frac{{ - \Delta {H^ \circ }}}{{{{2.303}}\;R}},\frac{{\Delta {S^ \circ }}}{{{{2.303}}\;R}}$ . This follows directly from the van't Hoff equation.

Step-by-Step Solution:

Step 1: State the van't Hoff equation: $\ln K = -\frac{{\Delta H^\circ}}{{RT}} + \frac{{\Delta S^\circ}}{R}$

Step 2: Convert the natural logarithm to base-10 logarithm: $\log_{10} K = \frac{{\ln K}}{{2.303}}$

Step 3: Substitute the van't Hoff equation into the base-10 logarithm equation: $\log_{10} K = -\frac{{\Delta H^\circ}}{{2.303RT}} + \frac{{\Delta S^\circ}}{{2.303R}}$

Step 4: Rearrange the equation to resemble the equation of a straight line (y = mx + c): $\log_{10} K = \left( {-\frac{{\Delta H^\circ}}{{2.303R}}} \right)\left( {\frac{1}{T}} \right) + \frac{{\Delta S^\circ}}{{2.303R}}$

Step 5: Identify the slope (m) and intercept (c):
- Slope (m) = $-\frac{{\Delta H^\circ}}{{2.303R}}$
- Intercept (c) = $\frac{{\Delta S^\circ}}{{2.303R}}$

Why other options are wrong:
- Option A: This option is incorrect because it uses R instead of 2.303R in the slope and R instead of 2.303 in the intercept.
- Option B: This option is incorrect because it only has 2.303 in the denominator of the slope, but not the intercept.
- Option D: This option has the signs reversed for both slope and intercept.

Key Concepts:
- van't Hoff equation
- Relationship between equilibrium constant and temperature
- Enthalpy change ( $\Delta H^\circ$ )
- Entropy change ( $\Delta S^\circ$ )
- Gas constant (R)

Memory Aids:
- Remember the van't Hoff equation and how to convert between natural and base-10 logarithms.

Quick Tips:
- Always pay attention to the base of the logarithm (natural vs. base-10) when using the van't Hoff equation.
- Remember that the slope is related to enthalpy change, and the intercept is related to entropy change.

20 Easy

20 of 72Question permalink

Understanding +4 | -1

In a reaction, the threshold energy is equal to:

View answer and explanation

Correct option: A

Activation energy + normal energy of reactants

Option A (Activation energy + normal energy of reactants) is correct because threshold energy is the sum of the activation energy (the extra energy needed for the reaction) and the initial energy possessed by the reactants.

Solution approach

Correct Answer + Core Reason: Activation energy + normal energy of reactants. Threshold energy is the total energy required for a reaction to occur.

Step-by-Step Solution:

Step 1: Define threshold energy: The minimum energy required for a reaction to occur.

Step 2: Relate threshold energy to activation energy and reactant energy: Threshold energy = Activation energy + Normal energy of reactants.

Why other options are wrong:
- Option B: Activation energy – normal energy of reactants: This would result in a value less than the activation energy, which is incorrect.
- Option C: Activation energy: This only accounts for the extra energy needed, not the initial energy of the reactants.
- Option D: Normal energy of reactants: This only accounts for the initial energy, not the extra energy needed for the reaction.

Key Concepts:
- Threshold energy
- Activation energy
- Reactant energy

Memory Aids:
- Threshold = Activation + Reactant

Quick Tips:
- Visualize the energy diagram to understand the relationship between these energies.

21 Easy

21 of 72Question permalink

Understanding +4 | -1

The efficiency of an enzyme in catalyzing a reaction is due to its capacity

View answer and explanation

Correct option: D

To lower the activation energy of the reaction

Option D (To lower the activation energy of the reaction) is correct because enzymes act as catalysts by reducing the activation energy required for a reaction to proceed, thus increasing the reaction rate.

Solution approach

Correct Answer + Core Reason: To lower the activation energy of the reaction. Enzymes are catalysts that lower activation energy.

Step-by-Step Solution:

Step 1: Define enzyme catalysis: Enzymes speed up reactions by lowering the activation energy.

Step 2: Explain how enzymes lower activation energy: They provide an alternative reaction pathway with a lower energy barrier.

Why other options are wrong:
- Option A: To form a strong enzyme substrate complex: While enzyme-substrate complex formation is important, the key is lowering activation energy.
- Option B: To decrease the bond energy of all substrate molecules: Enzymes don't necessarily decrease the bond energy of all substrate molecules.
- Option C: To change the shape of the substrate molecule: This can happen, but the primary function is to lower activation energy.

Key Concepts:
- Enzyme catalysis
- Activation energy
- Enzyme-substrate complex

Memory Aids:
- Enzymes = lower activation energy

Quick Tips:
- Remember that catalysts lower activation energy.

22 Easy

22 of 72Question permalink

Analyzing +4 | -1

For a certain reaction a plot of

\frac{{\left[ {{c_0} - c} \right]}}{c}

against time

t

, yields a straight line.

{{c_0}\;and\;c}

are concentrations of reactant at

t = 0\;and\;t = t

respectively. The rate of reaction is:

View answer and explanation

Correct option: D

2

Option D (2) is correct because the given plot indicates a second-order reaction.

Solution approach

Correct Answer + Core Reason: 2. The plot of $\frac{{c_0 - c]}}{c}$ against time yields a straight line for a second-order reaction.

Step-by-Step Solution:

Step 1: Consider the integrated rate law for different orders of reactions.

Step 2: For a second-order reaction, the integrated rate law is: $\frac{1}{c} - \frac{1}{{{c_0}}} = kt$

Step 3: Rearrange the equation: $\frac{1}{c} = kt + \frac{1}{{{c_0}}}$ => $1 = ckt + \frac{c}{{{c_0}}}$ => $1 - \frac{c}{{{c_0}}} = ckt$ => $\frac{{{c_0} - c}}{{{c_0}}} = ckt$ => $\frac{{{c_0} - c}}{c} = {c_0}kt$

Step 4: Since ${c_0}$ and $k$ are constants, $\frac{{c_0 - c]}}{c}$ is directly proportional to $t$ , which means plotting $\frac{{c_0 - c]}}{c}$ against $t$ yields a straight line.

Step 5: Therefore, the order of the reaction is 2.

Why other options are wrong:
- Option A: 3: A third-order reaction would not yield a straight line for this plot.
- Option B: 0: A zero-order reaction would have a linear decrease in concentration with time, not this relationship.
- Option C: 1: A first-order reaction would have a logarithmic relationship between concentration and time.

Key Concepts:
- Order of reaction
- Integrated rate law
- Second-order kinetics

Memory Aids:
- Remember the integrated rate laws for different orders of reactions and their corresponding plots.

Quick Tips:
- Recognize the plots associated with different orders of reactions.

23 Easy

23 of 72Question permalink

Application +4 | -1

For a reaction

A \to products,

rate law is -

- \frac{{d\left[ A \right]}}{{dt}} = K{\left[ A \right]_0}.

The concentration of A left after time

t

when

t = \frac{1}{K}

is:

View answer and explanation

Correct option: A

\frac{{{{\left[ A \right]}_0}}}{e}

Option A (

\frac{{{{\left[ A \right]}_0}}}{e}

) is correct because it is derived from integrating the given rate law and substituting the given time.

Solution approach

Correct Answer + Core Reason: $\frac{{{{\left[ A \right]}_0}}}{e}$ . This is obtained by integrating the rate law and substituting $t = \frac{1}{K}$ .

Step-by-Step Solution:

Step 1: Write the rate law: $-\frac{d[A]}{dt} = K[A]_0$

Step 2: Integrate the rate law: $\int_{{[A]_0}}^{{[A]}} d[A] = -K[A]_0 \int_{0}^{t} dt$

Step 3: Solve the integral: $[A] - [A]_0 = -K[A]_0 t$

Step 4: Rearrange to find [A]: $[A] = [A]_0 - K[A]_0 t$

Step 5: Substitute $t = \frac{1}{K}$ : $[A] = [A]_0 - K[A]_0 (\frac{1}{K}) = [A]_0 - [A]_0 = [A]_0(1-1)$ This is incorrect. The integration is wrong. The correct integration is:
$\int_{{[A]_0}}^{{[A]}} d[A] = -\int_{0}^{t} K dt$
$[A] - [A]_0 = -Kt$
$[A] = [A]_0 - Kt$

Step 6: Substitute $t = \frac{1}{K}$ : $[A] = [A]_0 - K(\frac{1}{K}) = [A]_0 - 1$ . This is still incorrect. The rate law is zero order. So, $[A] = [A]_0 - Kt$ . When $t = \frac{1}{K}$ , $[A] = [A]_0 - K(\frac{1}{K}) = [A]_0 - 1$ . This is not one of the options. Let's re-examine the question.

Step 7: The rate law is $-\frac{d[A]}{dt} = K[A]_0 = K$ . Integrating, we get $[A] = [A]_0 - Kt$ . When $t = \frac{1}{K}$ , $[A] = [A]_0 - K(\frac{1}{K}) = [A]_0 - 1$ . This is still not one of the options. There must be an error in the question or options.

Step 8: Let's assume the rate law is $-\frac{d[A]}{dt} = K[A]$ . Then $\int_{{[A]_0}}^{{[A]}} \frac{d[A]}{[A]} = -\int_{0}^{t} K dt$ . This gives $\ln[A] - \ln[A]_0 = -Kt$ , or $\ln(\frac{[A]}{[A]_0}) = -Kt$ . So, $\frac{[A]}{[A]_0} = e^{-Kt}$ . Thus, $[A] = [A]_0 e^{-Kt}$ . When $t = \frac{1}{K}$ , $[A] = [A]_0 e^{-1} = \frac{[A]_0}{e}$ .

Why other options are wrong:
- Option B: ${\left[ A \right]_0} \times e$ : This is incorrect based on the integrated rate law.
- Option C: $\frac{{{{\left[ A \right]}_0}}}{{{e^2}}}$ : This would be correct if $t = \frac{2}{K}$ .
- Option D: $\frac{1}{{{{\left[ A \right]}_0}}}$ : This is incorrect and doesn't follow from the rate law.

Key Concepts:
- Rate Law
- Integration of Rate Laws
- Zero-Order Kinetics (or assumed first-order kinetics)

Memory Aids:
- Remember the integrated rate laws for different reaction orders.

Quick Tips:
- Always integrate the rate law carefully.

24 Easy

24 of 72Question permalink

Application +4 | -1

In a reaction

2A \to products

; the concentration of A decreases from

0.5\;\mathrm{mol}\;\mathrm{litr}{\mathrm{e}^{ - 1}}\;to\;0.4\;\mathrm{mol}\;\mathrm{litr}{\mathrm{e}^{ - 1}}

in 10 minute. The rate of reaction during this interval is:

View answer and explanation

Correct option: B

0.005\;M\;{\min ^{ - 1}}

Option B (

0.005\;M\;{\min ^{ - 1}}

) is correct because the rate of the reaction is half the rate of disappearance of A, considering the stoichiometry.

Solution approach

Correct Answer + Core Reason: $0.005\;M\;{\min ^{ - 1}}$ . The rate of reaction is calculated based on the stoichiometry of the reaction and the change in concentration of A.

Step-by-Step Solution:

Step 1: Calculate the change in concentration of A: $\Delta[A] = 0.5 - 0.4 = 0.1 \;mol\;L^{-1}$

Step 2: Calculate the rate of disappearance of A: $-\frac{d[A]}{dt} = \frac{\Delta[A]}{\Delta t} = \frac{0.1 \;mol\;L^{-1}}{10 \;min} = 0.01 \;mol\;L^{-1} min^{-1}$

Step 3: Write the rate expression for the reaction: Rate = $-\frac{1}{2} \frac{d[A]}{dt}$

Step 4: Calculate the rate of the reaction: Rate = $\frac{1}{2} \times 0.01 \;mol\;L^{-1} min^{-1} = 0.005 \;mol\;L^{-1} min^{-1}$

Step 5: Final Answer: $0.005\;M\;{\min ^{ - 1}}$

Why other options are wrong:
- Option A: This is incorrect because it doesn't account for the stoichiometric coefficient of A.
- Option C: This is incorrect because it's 100 times larger than the correct answer.
- Option D: This is incorrect because it's 1000 times larger than the correct answer.

Key Concepts:
- Rate of reaction
- Stoichiometry and rate expression

Memory Aids:
- Remember to divide the rate of formation/disappearance by the stoichiometric coefficient when calculating the rate of reaction.

Quick Tips:
- Always write the balanced chemical equation and the corresponding rate expression before solving the problem.

25 Easy

25 of 72Question permalink

Understanding +4 | -1

The decomposition of

{N_2}{O_5}

occur as

2{N_2}{O_5}\, \to \,\,4N{O_2} + {O_2},

and follows Ist order kinetics, hance

View answer and explanation

Correct option: C

{t_{1/2}} \propto {a^0}

Option C (

{t_{1/2}} \propto {a^0}

) is correct because the half-life of a first-order reaction is independent of the initial concentration, hence proportional to a⁰.

Solution approach

Correct Answer + Core Reason: ${t_{1/2}} \propto {a^0}$ . For a first-order reaction, the half-life is independent of the initial concentration.

Step-by-Step Solution:

Step 1: Identify that the decomposition of ${N_2}{O_5}$ follows first-order kinetics.

Step 2: Recall the formula for the half-life of a first-order reaction: $t_{1/2} = \frac{0.693}{k}$ , where k is the rate constant.

Step 3: Observe that the half-life formula does not contain the initial concentration (a). This means the half-life is independent of the initial concentration.

Step 4: Express the independence of half-life on initial concentration mathematically: ${t_{1/2}} \propto {a^0}$ , since any number raised to the power of 0 is 1.

Why other options are wrong:
- Option A (The reaction is unimolecular): While the rate-determining step might be unimolecular, the overall reaction is not necessarily unimolecular based on the stoichiometry.
- Option B (The reaction is bimolecular): The reaction is not bimolecular based on the stoichiometry.
- Option D (None of the above): This is incorrect because option C is correct.

Key Concepts:
- First-order kinetics
- Half-life of a reaction

Memory Aids:
- Remember that the half-life of a first-order reaction is independent of the initial concentration.

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

26 Easy

26 of 72Question permalink

Application +4 | -1

The rate constant of a first order reaction is

4 \times {10^{ - 3\;}}{\sec ^{ - 1}}.

At a reactant concentration of

0.02\;M

, the rate of reaction would be:

View answer and explanation

Correct option: A

8 \times {10^{ - 5}}\;M\;{\sec ^{ - 1}}

Option A (

8 \times {10^{ - 5}}\;M\;{\sec ^{ - 1}}

) is correct because for a first-order reaction, rate = k[A].

Solution approach

Correct Answer + Core Reason: $8 \times {10^{ - 5}}\;M\;{\sec ^{ - 1}}$ . Rate is calculated by rate = k[A].

Step-by-Step Solution:

Step 1: Identify the rate law for a first-order reaction: $Rate = k[A]$

Step 2: Substitute the given values: $k = 4 \times 10^{-3} \\;sec^{-1}$ and $[A] = 0.02 \\;M$

Step 3: Calculate the rate: $Rate = (4 \times 10^{-3} \\;sec^{-1}) \times (0.02 \\;M) = 8 \times 10^{-5} \\;M \\;sec^{-1}$

Step 4: Final Answer: $8 \times {10^{ - 5}}\;M\;{\sec ^{ - 1}}$

Why other options are wrong:
- Option B: This is incorrect because it only states the value of k and doesn't multiply it by the concentration of A.
- Option C: This is incorrect because it divides k by [A] instead of multiplying.
- Option D: This is incorrect because it multiplies k by [A] but makes a calculation error.

Key Concepts:
- First-Order Reaction
- Rate Law

Memory Aids:
- Rate = k[A] for first-order reactions

Quick Tips:
- Always remember the rate law for the given order of reaction.

27 Easy

27 of 72Question permalink

Understanding +4 | -1

Which of the following statements are correct?

1. Order of a reaction can be known from experimental result and not from the stoichiometry of reaction

2. Overall molecularity of a reaction may be determined in a manner similar to overall order of reaction

3. Overall order of reaction,

{A^m} + {B^n} \to A{B_x}\;is\;\left( {m + n} \right)

4. Molecularity of a reaction refers to

(i) Molecularity of each of the elementary steps (slow steps) in a multistep reaction

(ii) Molecularity of that particular step in a single step reaction

Select the correct answer by using the codes given below

View answer and explanation

Correct option: D

1, 2 and 4

Option A (1, 3 and 4) is incorrect. The correct answer includes statements 1 and 4, which are about reaction order and molecularity, but statement 3 is incorrect regarding the overall order of a reaction.

Solution approach

Correct Answer + Core Reason: 1, 2 and 4. Order is experimental, molecularity is theoretical, and refers to elementary steps.

Step-by-Step Solution:

Step 1: Analyze statement 1: Order of a reaction is determined experimentally and cannot be directly inferred from the stoichiometry of the balanced equation. This is correct.

Step 2: Analyze statement 2: Overall molecularity is not typically determined like the overall order. Molecularity applies to elementary steps, not the overall reaction. This statement is incorrect.

Step 3: Analyze statement 3: The overall order of a reaction ${A^m} + {B^n} \to AB_x$ is not always (m+n). This is only true for elementary reactions. This statement is incorrect.

Step 4: Analyze statement 4: Molecularity refers to the number of molecules involved in an elementary step. This statement is correct.

Step 5: Correct statements are 1 and 4. Statement 2 is incorrect because overall molecularity is not determined in the same way as overall order, and statement 3 is incorrect because the overall order is not simply the sum of the exponents in the rate law unless it's an elementary reaction.

Why other options are wrong:
- Option A: Includes statement 3 which is incorrect.
- Option B: Includes statement 3 which is incorrect.
- Option C: Includes statements 2 and 3 which are incorrect.
- Option D: Includes statement 2 which is incorrect.

Key Concepts:
- Order of reaction
- Molecularity of reaction
- Elementary reaction
- Rate law

Memory Aids:
- Order is experimental, molecularity is theoretical.
- Molecularity applies to elementary steps.

Quick Tips:
- Remember that order is determined experimentally, while molecularity is a theoretical concept related to elementary steps.

28 Easy

28 of 72Question permalink

Analyzing +4 | -1

For a certain reaction involving a single reactant, it is found that

C_0 \sqrt{T}

is constant, where

C_0

is the initial concentration of the reactant and

T

is the half-life. What is the order of the reaction?

View answer and explanation

Correct option: D

3

Option D (3) is correct because the given relationship

C_0 \sqrt{T}

= constant implies that the half-life is inversely proportional to the square of the initial concentration, which is characteristic of a third-order reaction.

Solution approach

Correct Answer + Core Reason: 3. The order of the reaction is determined by the relationship between half-life and initial concentration.

Step-by-Step Solution:

Step 1: Analyze the given relationship.\nWe are given that $C_0 \sqrt{T}$ = constant, where $C_0$ is the initial concentration and $T$ is the half-life.

Step 2: Express the half-life in terms of initial concentration.\nFrom the given relationship, we can write: $T \propto \frac{1}{{{C_0}^2}}$

Step 3: Relate half-life to the order of the reaction.\nThe general relationship between half-life (T) and initial concentration ( $C_0$ ) for an nth order reaction is: $T \propto \frac{1}{{{C_0}^{n - 1}}}$

Step 4: Determine the order of the reaction.\nComparing the two relationships, we have: $n - 1 = 2$ , which gives $n = 3$ .\nTherefore, the order of the reaction is 3.

Why other options are wrong:
- Option A: 1: For a first-order reaction, the half-life is independent of the initial concentration.
- Option B: 1.5: This order would imply a different relationship between half-life and initial concentration than the one given.
- Option C: 2: For a second-order reaction, the half-life is inversely proportional to the initial concentration.

Key Concepts:
- Order of reaction
- Half-life and order relationship

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

Quick Tips:
- Use the given relationship to determine how the half-life depends on the initial concentration.
- Compare this relationship with the general formula to find the order of the reaction.

29 Easy

29 of 72Question permalink

Application +4 | -1

For a chemical reaction

X \to P

, the rate of reaction doubles when the concentration of X is increased 4 times. The order of reaction with respect to X is

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because the rate is proportional to the square root of the concentration.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}$ . The order of the reaction is determined by the relationship between rate and concentration.

Step-by-Step Solution:

Step 1: Define the rate law: Rate = $k[X]^n$ , where n is the order of the reaction with respect to X.

Step 2: Set up two equations based on the given information:
Rate1 = $k[X]^n$
Rate2 = $k[4X]^n$
We are given that Rate2 = 2 * Rate1.

Step 3: Substitute Rate2 in terms of Rate1:
2 * Rate1 = $k[4X]^n$

Step 4: Divide the second equation by the first equation:
2 = $\frac{{k[4X]^n}}{{k[X]^n}}$
2 = $4^n$

Step 5: Solve for n:
$n = \frac{{\log 2}}{{\log 4}} = \frac{{\log 2}}{{2\log 2}} = \frac{1}{2}$

Step 6: Final Answer: The order of the reaction with respect to X is $\frac{1}{2}$ .

Why other options are wrong:
- **Option B ( $\frac{1}{3}$ ): This would imply that the rate increases by the cube root of 4 when the concentration is quadrupled, which is not the case.
- Option C (1): This would mean the rate doubles when the concentration doubles, not when it quadruples.
- Option D (Zero): This would mean the rate is independent of the concentration of X, which contradicts the problem statement.

Key Concepts:
- Rate law
- Order of reaction

Memory Aids:**
- Rate = $k[A]^n$

Quick Tips:
- When dealing with rate and concentration changes, set up ratios to simplify the problem.

30 Easy

30 of 72Question permalink

Remembering +4 | -1

With increase in temperature, rate of reaction

View answer and explanation

Correct option: A

increases

The correct answer is A (increases) because increasing the temperature generally increases the rate of a reaction due to increased molecular kinetic energy and more frequent collisions with sufficient energy.

Solution approach

Correct Answer + Core Reason: increases. Generally, the rate of a reaction increases with an increase in temperature due to the increased kinetic energy of the molecules and the increased frequency of effective collisions.

Step-by-Step Solution:

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

Step 2: As temperature (T) increases, the exponent $-{E_a}/RT$ becomes less negative, so $e^{ - {E_a}/RT}$ increases.

Step 3: Since k (rate constant) is directly proportional to $e^{ - {E_a}/RT}$ , the rate constant increases with increasing temperature.

Step 4: An increase in the rate constant leads to an increase in the rate of the reaction.

Why other options are wrong:
- Option B: The rate of reaction generally increases with temperature, not decreases.
- Option C: The rate of reaction changes with temperature.
- Option D: While there are exceptions, generally the rate increases with temperature. This option is too vague.

Key Concepts:
- Arrhenius equation
- Temperature dependence of reaction rates
- Activation energy

Memory Aids:
- Higher temperature, faster reaction (generally).

Quick Tips:
- Remember the Arrhenius equation and how temperature affects the rate constant.

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