Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Chemical Kinetics

Question:

Read the passage carefully and answer the questions based on the passage:

The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. In the rate equation Rate = $k [A]^x [B]^y$

$x$ and $y$ indicate how sensitive the rate is to the change in concentration of A and B, respectively. Sum of these exponents, i.e., $x + y$ gives the overall order of a reaction where $x$ and $y$ represent the order with respect to the reactants A and B, respectively. Hence, the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. For a first order reaction, the concentration of the reactant varies as $[R] = [R]_0e^{-kt}$

The initial concentration of N2O5 in the following first order reaction $N_2O_5(g) → 2 NO_2(g) + 1/2O_2 (g)$ was $1.2 × 10^{-2}\, mol\, L^{-1}$ at 298 K. The concentration of $N_2O_5$ after 30 minutes was $0.60 × 10^{-2}\, mol\, L^{-1}$. Calculate the rate constant of the reaction at 298 K.

Options:

$0.040\, min^{-1}$

$0.023\, min^{-1}$

$0.032\, min^{-1}$

$0.050\, min^{-1}$

Correct Answer:

$0.023\, min^{-1}$

Explanation:

The correct answer is Option (2) → $0.023\, min^{-1}$

The reaction is a first-order reaction, so you can use the integrated rate law to calculate the rate constant ($k$):

$k = \frac{2.303}{t} \log \frac{[\text{A}]_0}{[\text{A}]_t}$

Where:

  • $t = 30\ \text{minutes}$
  • $[\text{A}]_0$ (initial concentration of $\text{N}_2\text{O}_5$) $= 1.2 \times 10^{-2}\ \text{mol}\ \text{L}^{-1}$
  • $[\text{A}]_t$ (concentration of $\text{N}_2\text{O}_5$ at time $t$) $= 0.60 \times 10^{-2}\ \text{mol}\ \text{L}^{-1}$

Calculation

Substitute the given values into the equation:

$k = \frac{2.303}{30\ \text{min}} \log \frac{1.2 \times 10^{-2}}{0.60 \times 10^{-2}}$

Simplify the ratio inside the logarithm:

$k = \frac{2.303}{30\ \text{min}} \log 2$

Calculate the value (using $\log 2 \approx 0.3010$):

$k = \frac{2.303 \times 0.3010}{30\ \text{min}}$

$k = \frac{0.6931}{30\ \text{min}}$

$k \approx 0.0231\ \text{min}^{-1}$

The rate constant is approximately $0.023\ \text{min}^{-1}$.