Question

At \(T(\mathrm{K})\), decomposition of \(A(g)\) follows first order kinetics.

\[ A(g) \rightarrow B(g) + C(g) \]
The following data is obtained for this reaction.

\[ \begin{array}{|c|c|} \hline \text{Time (s)} & \text{Total pressure (atm)} \\ \hline 0 & 0.5 \\ \hline 100 & 0.6 \\ \hline \end{array} \]
What is the rate constant (in \(\mathrm{s^{-1}}\)) for this reaction?

\[ \left(\log 1.25 = 0.097,\; \log 1.666 = 0.222\right) \]

• A. \(2.23\times10^{-3}\)
• B. \(5.11\times10^{-3}\)
• C. \(2.23\times10^{-4}\)
• D. \(5.11\times10^{-4}\)

Hint:

For gaseous first-order reactions, \[ \boxed{ k=\frac{2.303}{t}\log\left(\frac{P_0}{P_t}\right) } \] where \(P_t\) is the partial pressure of the reactant at time \(t\), not the total pressure.

Answer 1

The Correct Option is A

Step 1: Determine the partial pressure of \(A\) after \(100\) s. Initially, \[ P_0=0.5\ \text{atm} \] Let the pressure decrease due to decomposition be \(x\). Then, \[ A \rightarrow B+C \] \[ \begin{array}{c|ccc} & A & B & C \hline \text{Initially} & 0.5 & 0 & 0 \text{After time }t & 0.5-x & x & x \end{array} \] Total pressure after \(100\) s: \[ (0.5-x)+x+x=0.5+x=0.6 \] \[ x=0.1\ \text{atm} \] Hence, \[ P_A=0.5-0.1=0.4\ \text{atm} \]

Step 2: Apply the first-order rate equation. \[ k=\frac{2.303}{t}\log\left(\frac{P_0}{P_A}\right) \] Substituting, \[ k=\frac{2.303}{100}\log\left(\frac{0.5}{0.4}\right) \] \[ =\frac{2.303}{100}\log(1.25) \] Given, \[ \log(1.25)=0.097 \] Therefore, \[ k=\frac{2.303\times0.097}{100} =2.23\times10^{-3}\ \mathrm{s^{-1}} \] Thus, \[ \boxed{k=2.23\times10^{-3}\ \mathrm{s^{-1}}} \] Hence, \[ \boxed{(A)} \] is the correct answer.