Question

At \(T\)(K), the following data is obtained for the decomposition of \(A(g)\), which follows first order kinetics \[ A(g)\rightarrow B(g)+C(g) \] \[ \begin{array}{|c|c|} \hline \text{Time (s)} & \text{Total pressure (atm)} \hline 0 & 0.5 100 & 0.6 x & 0.65 \hline \end{array} \] What is \(x\) (in s)? \[ (\log1.25=0.097,\qquad \log1.4285=0.1549) \]

• A. \(180\)
• B. \(160\)
• C. \(140\)
• D. \(200\)

Hint:

For the decomposition \[ \boxed{A\rightarrow B+C,} \] the partial pressure of reactant is \[ \boxed{ P_A=2P_0-P_t. } \] Then use the first-order equation \[ \boxed{ k=\frac{2.303}{t}\log\frac{P_0}{P_A}. } \]

Answer 1

The Correct Option is B

Step 1: Determine the partial pressure of \(A\). Initially, \[ P_0=0.5\text{ atm}. \] For the reaction \[ A\rightarrow B+C, \] the partial pressure of \(A\) at any instant is \[ P_A=2P_0-P_t. \] At \(t=100\) s, \[ P_A=2(0.5)-0.6=0.4\text{ atm}. \] At \(t=x\), \[ P_A=2(0.5)-0.65=0.35\text{ atm}. \]

Step 2: Use the first-order rate equation. For a first-order reaction, \[ k=\frac{2.303}{t}\log\frac{P_0}{P_A}. \] Using \(t=100\) s, \[ k = \frac{2.303}{100}\log\frac{0.5}{0.4} = \frac{2.303}{100}(0.097). \] Again, \[ k = \frac{2.303}{x}\log\frac{0.5}{0.35} = \frac{2.303}{x}(0.1549). \]

Step 3: Calculate \(x\). Equating the two expressions, \[ \frac{0.097}{100} = \frac{0.1549}{x}. \] Therefore, \[ x = 100\left(\frac{0.1549}{0.097}\right) \approx160\text{ s}. \] Hence, \[ \boxed{160\text{ s}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.