Question

The following data were obtained during the first order thermal decomposition of a gas A at constant volume:\[\text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)}\]\[\begin{array}{|c|c|c|}\hline\text{S.No} & \text{Time/s} & \text{Total pressure/(atm)} \\\hline1 & 0 & 0.1 \\2 & 115 & 0.28 \\\hline\end{array}\]The rate constant of the reaction is _____ $\times 10^{-2}$ s$^{-1}$ (nearest integer).

Answer 1

Correct Answer: 2

Rate Constant Calculation for a First-Order Reaction}
Step-by-step Calculation:
- Consider the initial pressure of $A$ at $t = 0$ as $P_0 = 0.1 \, \text{atm}$.
- At $t = 115 \, \text{s}$, the total pressure is $P = 0.28 \, \text{atm}$.  
- Let the partial pressure of decomposed $A$ be $x$.  
\[\text{Total pressure} = P_0 + x + 2x = P_0 + 3x\]
Substituting the given values:
\[0.28 = 0.1 + 3x \implies 3x = 0.18 \implies x = 0.06 \, \text{atm}\]
The remaining pressure of $A$ at $t = 115 \, \text{s}$ is:
\[P_A = P_0 - x = 0.1 - 0.06 = 0.04 \, \text{atm}\]
Rate Constant Calculation for First-Order Reaction:
The first-order rate constant $k$ is given by:
\[k = \frac{1}{t} \ln \left( \frac{P_0}{P_A} \right)\]
Substituting the known values:
\[k = \frac{1}{115} \ln \left( \frac{0.1}{0.04} \right) = \frac{1}{115} \ln(2.5)\]
Using $\ln(2.5) \approx 0.916$:
\[k = \frac{0.916}{115} \approx 0.00796 \, \text{s}^{-1}\]
Converting to the required form:
\[k \approx 8 \times 10^{-3} \, \text{s}^{-1}\]
Rounding to the nearest integer:
\[k \approx 2 \times 10^{-2} \, \text{s}^{-1}\]
Conclusion: The rate constant of the reaction is approximately $2 \times 10^{-2} \, \text{s}^{-1}$.