Question

Consider a general first order reaction \(\text{A(g)} \rightarrow \text{B(g)} + \text{C(g)}\). If the initial pressure is 200 mm and after 20 minutes it is 250 mm, then the half-life period of the reaction (in minutes) is
(\(\log 2 = 0.30, \log 3 = 0.48, \log 4 = 0.60\))

• A. 40.2
• B. 50.2
• C. 20.5
• D. 60.5

Hint:

For decomposition of gas A into n gaseous products (\(A \rightarrow n Products\)): \(k = \frac{2.303}{t} \log \left( \frac{(n-1)P_0}{nP_0 - P_t} \right)\). Here \(n=2\), so \(k \propto \log \frac{P_0}{2P_0 - P_t}\).

Answer 1

The Correct Option is B

Step 1: Express Pressures at time t:
\[ \text{A(g)} \rightarrow \text{B(g)} + \text{C(g)} \] Initial (\(t=0\)): \(P_0\) \quad 0 \quad 0 At time \(t\): \(P_0 - p\) \quad \(p\) \quad \(p\) Total Pressure \(P_t = (P_0 - p) + p + p = P_0 + p\). From this, \(p = P_t - P_0\). Pressure of reactant A remaining, \(P_A = P_0 - p = P_0 - (P_t - P_0) = 2P_0 - P_t\).
Step 2: Substitute Values:
Given \(P_0 = 200 \text{ mm}\), \(P_t = 250 \text{ mm}\) at \(t = 20 \text{ min}\). \(P_A = 2(200) - 250 = 400 - 250 = 150 \text{ mm}\).
Step 3: Calculate Rate Constant (k):
For first order: \(k = \frac{2.303}{t} \log \left( \frac{P_0}{P_A} \right)\) \(k = \frac{2.303}{20} \log \left( \frac{200}{150} \right) = \frac{2.303}{20} \log \left( \frac{4}{3} \right)\). Using given logs: \(\log 4 - \log 3 = 0.60 - 0.48 = 0.12\). \(k = \frac{2.303 \times 0.12}{20}\).
Step 4: Calculate Half-Life (\(t_{1/2}\)):
\(t_{1/2} = \frac{0.693}{k} = \frac{2.303 \times \log 2}{k}\). \(t_{1/2} = \frac{2.303 \times 0.30}{\frac{2.303 \times 0.12}{20}}\) \(t_{1/2} = \frac{0.30 \times 20}{0.12} = \frac{6}{0.12} = 50 \text{ min}\). Given the options, 50.2 is the closest and correct answer. Final Answer:
50.2 min.