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. |
$0.040\, min^{-1}$ $0.023\, min^{-1}$ $0.032\, min^{-1}$ $0.050\, min^{-1}$ |
$0.023\, min^{-1}$ |
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:
CalculationSubstitute 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 $0.023\ \text{min}^{-1}$. |