Chemical Kinetics Numerical Practice Sheet - Vardaan Learning Institute

Q1. Rate Expression CBSE 2019

For the reaction $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$, the rate of formation of $NO_2(g)$ is $2.8 \times 10^{-3} \text{ M s}^{-1}$. Calculate the rate of disappearance of $N_2O_5(g)$.

Solution:

$$ -\frac{1}{2}\frac{d[N_2O_5]}{dt} = +\frac{1}{4}\frac{d[NO_2]}{dt} $$

$$ -\frac{d[N_2O_5]}{dt} = \frac{2}{4} \times \left(2.8 \times 10^{-3}\right) = \mathbf{1.4 \times 10^{-3} \text{ M s}^{-1}} $$

Q2. Effect of Volume CBSE 2024

For a reaction $A + B \rightarrow P$, the rate law is given by $\text{Rate} = k[A][B]^2$. How is the rate of reaction affected if the volume of the reaction vessel is suddenly reduced to half of its original volume?

Solution:

When the volume is reduced to half ($V' = V/2$), the concentration of all gases doubles since $C = n/V$.

New concentrations: $[A]' = 2[A]$, $[B]' = 2[B]$.

New rate $r_2 = k[A]'([B]')^2 = k(2[A])(2[B])^2 = 8k[A][B]^2 = 8 r_1$. The rate increases to 8 times.

Q3. Disappearance vs Reaction Rate CBSE 2025

For the reaction $2A \rightarrow P$, the rate of disappearance of A is $0.05 \text{ M s}^{-1}$. Find the rate of the overall reaction and the rate of formation of P.

Solution:

Rate of reaction = $-\frac{1}{2}\frac{d[A]}{dt} = +\frac{d[P]}{dt}$

Given $-\frac{d[A]}{dt} = 0.05 \text{ M s}^{-1}$

Rate of reaction = $\frac{1}{2} \times 0.05 = \mathbf{0.025 \text{ M s}^{-1}}$

Rate of formation of P ($\frac{d[P]}{dt}$) = Rate of reaction = $\mathbf{0.025 \text{ M s}^{-1}}$

Q4. Complex Stoichiometry CBSE 2026

For the reaction $4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)$, the rate of formation of NO is $3.6 \times 10^{-3} \text{ M s}^{-1}$. Calculate the rate of disappearance of $NH_3$ and rate of formation of $H_2O$.

Solution:

Rate = $-\frac{1}{4}\frac{d[NH_3]}{dt} = +\frac{1}{4}\frac{d[NO]}{dt} = +\frac{1}{6}\frac{d[H_2O]}{dt}$

Since the coefficients of $NH_3$ and $NO$ are both 4, rate of disappearance of $NH_3$ = rate of formation of $NO$ = $\mathbf{3.6 \times 10^{-3} \text{ M s}^{-1}}$.

For $H_2O$: $\frac{1}{6}\frac{d[H_2O]}{dt} = \frac{1}{4}\frac{d[NO]}{dt}$

$\frac{d[H_2O]}{dt} = \frac{6}{4} \times 3.6 \times 10^{-3} = 1.5 \times 3.6 \times 10^{-3} = \mathbf{5.4 \times 10^{-3} \text{ M s}^{-1}}$.

Q5. Initial Rate Method CBSE 2025

From the following data for $A + B \rightarrow \text{Products}$, find the order of reaction w.r.t A and B.

Exp	[A] (mol/L)	[B] (mol/L)	Initial Rate (M/s)
1	0.10	0.10	$2.0 \times 10^{-3}$
2	0.10	0.20	$4.0 \times 10^{-3}$
3	0.20	0.10	$8.0 \times 10^{-3}$

Solution:

Let the rate law be $\text{Rate} = k[A]^x[B]^y$.

From Exp 1 & 2: $[A]$ is constant. Doubling $[B]$ doubles the rate. $\Rightarrow \mathbf{y = 1}$

From Exp 1 & 3: $[B]$ is constant. Doubling $[A]$ quadruples the rate. $\Rightarrow \mathbf{x = 2}$

Rate Law: $\text{Rate} = k[A]^2[B]^1$

Q6. Units of Rate Constant CBSE 2026

Identify the reaction order from each of the following rate constants:
(i) $k = 2.3 \times 10^{-5} \text{ L mol}^{-1} \text{ s}^{-1}$
(ii) $k = 3 \times 10^{-4} \text{ s}^{-1}$
(iii) $k = 4.5 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}$

Solution:

The unit of rate constant $k$ is: $(\text{mol L}^{-1})^{1-n} \text{ s}^{-1}$

(i) $\text{L mol}^{-1} \text{ s}^{-1} \Rightarrow 1 - n = -1 \Rightarrow \mathbf{n = 2}$ (Second Order)

(ii) $\text{s}^{-1} \Rightarrow 1 - n = 0 \Rightarrow \mathbf{n = 1}$ (First Order)

(iii) $\text{mol L}^{-1} \text{ s}^{-1} \Rightarrow 1 - n = 1 \Rightarrow \mathbf{n = 0}$ (Zero Order)

Q7. Fractional Order CBSE 2025

A reaction has the rate law: $\text{Rate} = k[A]^1[B]^{1/2}$. Calculate the overall order of the reaction. What will be the unit of the rate constant if concentration is in $\text{mol L}^{-1}$ and time in seconds?

Solution:

Overall order $n = 1 + \frac{1}{2} = \mathbf{\frac{3}{2}}$ or **1.5**.

Unit of $k = (\text{mol L}^{-1})^{1-n} \text{ s}^{-1}$

Unit $= (\text{mol L}^{-1})^{1 - 1.5} \text{ s}^{-1} = (\text{mol L}^{-1})^{-0.5} \text{ s}^{-1}$

Unit = $\mathbf{\text{L}^{1/2} \text{mol}^{-1/2} \text{ s}^{-1}}$

Q8. Concept of Rate Law CBSE 2026

The reaction $A + B \rightarrow C$ is zero order with respect to A and second order with respect to B. Write the rate law. How does the rate change if concentration of both A and B are doubled?

Solution:

Rate law: $\text{Rate} = k[A]^0[B]^2 = k[B]^2$

If concentrations of both A and B are doubled, $[A] \rightarrow 2[A]$ and $[B] \rightarrow 2[B]$.

New Rate = $k(2[B])^2 = 4k[B]^2$.

The rate increases to **4 times** its original value. (Changes in A have no effect because it is zero order w.r.t A).

Q9. Direct Calculation CBSE 2024

The rate constant for a first order reaction is $60 \text{ s}^{-1}$. How much time will it take to reduce the initial concentration of the reactant to its $1/16^{th}$ value?

Solution:

For a first order reaction: $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$

Given: $k = 60 \text{ s}^{-1}$, $[R] = \frac{[R]_0}{16}$.

$$ t = \frac{2.303}{60} \log (16) = \frac{2.303 \times 1.204}{60} = \mathbf{4.62 \times 10^{-2} \text{ s}} $$

Q10. Multi-Step Proof (99.9%) CBSE 2019, 2026

Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life ($t_{1/2}$) of the reaction.

Solution:

For 99.9% completion: $[R] = 0.1\%$ of $[R]_0$

$$ t_{99.9\%} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log 1000 = \frac{6.909}{k} $$

$$ t_{1/2} = \frac{0.693}{k} $$

$$ \frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909}{0.693} \approx 10 \Rightarrow \mathbf{t_{99.9\%} = 10 \times t_{1/2}} $$

Q11. Multi-Step Proof (99%) CBSE 2026

Show that the time required for 99% completion of a first-order reaction is twice the time required for 90% completion.

Solution:

For 99% completion: $[R] = 1\%$ of $[R]_0 = 1$ (if $[R]_0 = 100$).

$$ t_{99\%} = \frac{2.303}{k} \log \frac{100}{1} = \frac{2.303}{k} \log 100 = \frac{2.303 \times 2}{k} = \frac{4.606}{k} $$

For 90% completion: $[R] = 10\%$ of $[R]_0 = 10$.

$$ t_{90\%} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \log 10 = \frac{2.303 \times 1}{k} $$

Comparing the two: $\mathbf{t_{99\%} = 2 \times t_{90\%}}$ (Hence Proved).

Q12. Two Stage Kinetics CBSE 2025

A first order reaction takes 20 minutes for 25% decomposition. Calculate the time required for 75% decomposition. ($\log 3 = 0.4771$, $\log 4 = 0.6021$)

Solution:

Case 1 (25% decomp): $[R] = 75$, $t = 20 \text{ min}$

$$ k = \frac{2.303}{20} \log \frac{100}{75} = \frac{2.303}{20} \log \frac{4}{3} = \frac{2.303}{20} (0.6021 - 0.4771) = \frac{2.303 \times 0.125}{20} $$

$$ k = 0.01439 \text{ min}^{-1} $$

Case 2 (75% decomp): $[R] = 25$

$$ t = \frac{2.303}{k} \log \frac{100}{25} = \frac{2.303}{0.01439} \log 4 = 160.04 \times 0.6021 = \mathbf{96.36 \text{ min}} $$

Q13. Zero Order Basics CBSE 2025

A zero order reaction is 20% complete in 10 minutes. Calculate the time taken for 80% completion. Also state what the slope represents in the plot of $[R]$ vs $t$.

Solution:

For a zero-order reaction: $k = \frac{[R]_0 - [R]}{t}$. Slope of $[R]$ vs $t$ is **$-k$**.

Case 1 (20% complete): $k = \frac{100 - 80}{10} = 2 \text{ mol L}^{-1}\text{min}^{-1}$

Case 2 (80% complete): $2 = \frac{100 - 20}{t} \Rightarrow t = \frac{80}{2} = \mathbf{40 \text{ minutes}}$

Q14. Radioactive Decay NCERT

The half-life for radioactive decay of $^{14}\text{C}$ is 5730 years. An artifact had only 80% of the $^{14}\text{C}$ found in a living tree. Estimate the age of the sample. ($\log 1.25 = 0.0969$)

Solution:

$$ k = \frac{0.693}{5730} \text{ yr}^{-1} $$

$$ t = \frac{2.303}{k} \log \frac{100}{80} = \frac{2.303 \times 5730}{0.693} \log 1.25 $$

$$ t = 19041.8 \times 0.0969 = \mathbf{1845 \text{ years}} $$

Q15. Zero Order Half Life CBSE 2026

For a zero order reaction, the initial concentration of reactant is $0.1 \text{ M}$ and rate constant is $0.003 \text{ M s}^{-1}$. Calculate the half-life of the reaction.

Solution:

For a zero order reaction, the half-life is given by:

$$ t_{1/2} = \frac{[R]_0}{2k} $$

Given $[R]_0 = 0.1 \text{ M}$ and $k = 0.003 \text{ M s}^{-1}$.

$$ t_{1/2} = \frac{0.1}{2 \times 0.003} = \frac{0.1}{0.006} = \mathbf{16.67 \text{ s}} $$

Q16. Decomposition of NH3 CBSE 2025

Decomposition of $NH_3$ on a platinum surface is a zero order reaction: $2NH_3 \rightarrow N_2 + 3H_2$. What are the rates of production of $N_2$ and $H_2$ if $k = 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$?

Solution:

For a zero order reaction, $\text{Rate} = k[NH_3]^0 = k = 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$.

From stoichiometry: Rate = $-\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt}$

Rate of production of $N_2$ ($\frac{d[N_2]}{dt}$) = Rate = $\mathbf{2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}}$.

Rate of production of $H_2$ ($\frac{d[H_2]}{dt}$) = $3 \times \text{Rate} = 3 \times (2.5 \times 10^{-4}) = \mathbf{7.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}}$.

Q17. Temperature Coefficient CBSE 2026

The rate constant of a reaction increases by 5% when its temperature is raised from 27°C to 28°C. Calculate the activation energy of the reaction. ($R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}$)

Solution:

Given: $T_1 = 300\text{K}$, $T_2 = 301\text{K}$, $\frac{k_2}{k_1} = 1.05$.

$$ \log(1.05) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{301 - 300}{300 \times 301} \right] $$

$$ 0.0212 = \frac{E_a}{19.147} \left[ \frac{1}{90300} \right] \Rightarrow \mathbf{E_a = 36.65 \text{ kJ mol}^{-1}} $$

Q18. Fraction of Molecules CBSE 2024

The activation energy for the reaction $2HI(g) \rightarrow H_2(g) + I_2(g)$ is $209.5 \text{ kJ mol}^{-1}$ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

Solution:

Fraction of molecules $x = e^{-E_a / RT}$. Taking log: $\log x = -\frac{E_a}{2.303 RT}$

$$ \log x = -\frac{209500}{2.303 \times 8.314 \times 581} = -18.832 $$

$$ x = \text{Antilog} (-18.832) = \mathbf{1.47 \times 10^{-19}} $$

Q19. Frequency Factor CBSE 2026

The activation energy of a reaction is $50 \text{ kJ/mol}$ and the value of rate constant at $300 \text{ K}$ is $1.5 \times 10^{-4} \text{ s}^{-1}$. Calculate the frequency factor $A$. ($\text{Antilog } (3.834) = 6823$)

Solution:

According to Arrhenius Equation: $\log k = \log A - \frac{E_a}{2.303 RT}$

$$ \log(1.5 \times 10^{-4}) = \log A - \frac{50000}{2.303 \times 8.314 \times 300} $$

$$ \log 1.5 - 4 = \log A - 8.71 $$

$$ 0.176 - 4 = \log A - 8.71 \Rightarrow -3.824 = \log A - 8.71 $$

$$ \log A = 8.71 - 3.824 = 4.886 $$

$$ A = \text{Antilog}(4.886) = \mathbf{7.69 \times 10^4 \text{ s}^{-1}} $$

Q20. Rule of Thumb CBSE 2025

The rate of a reaction triples when the temperature changes from 20°C to 50°C. Calculate energy of activation. (Given $\log 3 = 0.4771$)

Solution:

Given: $T_1 = 20^\circ\text{C} = 293\text{K}$, $T_2 = 50^\circ\text{C} = 323\text{K}$, $\frac{k_2}{k_1} = 3$.

$$ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right] $$

$$ \log 3 = \frac{E_a}{2.303 \times 8.314} \left[ \frac{323 - 293}{293 \times 323} \right] $$

$$ 0.4771 = \frac{E_a}{19.147} \left[ \frac{30}{94639} \right] $$

$$ E_a = \frac{0.4771 \times 19.147 \times 94639}{30} = 28815 \text{ J mol}^{-1} $$

$$ \mathbf{E_a = 28.81 \text{ kJ mol}^{-1}} $$

Q21. Ester Hydrolysis CBSE 2024

Why is the hydrolysis of ethyl acetate in acidic medium a pseudo-first order reaction? State its rate law and what happens to the rate constant if the concentration of water is increased?

Solution:

Reaction: $\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$

Because water is present in such a large excess, its concentration remains almost constant. Therefore, the term $k'[\text{H}_2\text{O}]$ is taken as a new constant $k$.

New Rate Law: $\text{Rate} = k[\text{CH}_3\text{COOC}_2\text{H}_5]$, behaving as first-order.

Q22. Pressure Based Kinetics CBSE 2025

For the first order thermal decomposition $SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g)$ at constant volume:
At $t = 0$, $P_{total} = 0.5 \text{ atm}$.
At $t = 100 \text{ s}$, $P_{total} = 0.6 \text{ atm}$.
Calculate the rate constant of the reaction. ($\log(1.25) = 0.0969$)

Solution:

At time $t$, $P_A = 2P_i - P_t$

At $t = 100 \text{ s}$, $P_t = 0.6 \text{ atm}$. $P_A = 2(0.5) - 0.6 = 0.4 \text{ atm}$

$$ k = \frac{2.303}{100} \log \frac{0.5}{0.4} = \frac{2.303}{100} \log(1.25) = \frac{2.303 \times 0.0969}{100} = \mathbf{2.23 \times 10^{-3} \text{ s}^{-1}} $$

Q23. Inversion of Cane Sugar CBSE 2026

Inversion of cane sugar in acidic medium is pseudo first order. Write the chemical equation and state its rate law. Why is it pseudo-first order?

Solution:

Reaction: $\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{12}\text{O}_6 (\text{glucose}) + \text{C}_6\text{H}_{12}\text{O}_6 (\text{fructose})$

Rate Law: $\text{Rate} = k[\text{C}_{12}\text{H}_{22}\text{O}_{11}]$

It is pseudo-first order because although both cane sugar and water are reactants (making it theoretically a second-order reaction), water is present in huge excess. Its concentration does not change significantly during the reaction, making the rate dependent only on the concentration of cane sugar.

Q24. Decomposition Ratios CBSE 2025

The thermal decomposition of a compound is of first order. If 50% of a sample decomposes in 120 minutes, how much time will it take for 90% to decompose?

Solution:

Given 50% decomposition is $t_{1/2} = 120 \text{ min}$.

$$ k = \frac{0.693}{120} = 0.005775 \text{ min}^{-1} $$

For 90% decomposition: $[R] = 100 - 90 = 10$

$$ t_{90\%} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{0.005775} \log(10) $$

$$ t_{90\%} = \frac{2.303}{0.005775} = \mathbf{398.8 \text{ min}} $$

VARDAAN

Extended Numerical Practice Sheet: Chemical Kinetics

Type 1: Rate of Reaction & Stoichiometry

Type 2: Rate Law & Order of Reaction

Type 3: First Order Kinetics

Type 4: Zero Order Kinetics & Half Life

Type 5: Arrhenius Equation & Activation Energy

Type 6: Pseudo First Order & Pressure Based