Question

Consider the following first-order gas phase reaction at constant temperature \[ \text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)} \] If the total pressure of the gases is found to be 200 torr after 23 sec. and 300 torr upon the complete decomposition of \( \text{A} \) after a very long time, then the rate constant of the given reaction is \( \dots \dots \times 10^{-2} \, \text{s}^{-1} \) (nearest integer). \[ \text{[Given: } \log_{10}(2) = 0.301 \text{]} \]

Answer 1

Correct Answer: 3

To solve the problem, we consider the first-order reaction \(\text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)}\) and apply the concept of partial pressures in relation to time and constant temperature. The given data involves the pressures at two different times: 200 torr after 23 seconds and 300 torr at completion (long time).
The reaction involves a change in total pressure due to decomposition, initially high with unreacted \( \text{A} \), dropping as the products form.
Let's denote the initial pressure of \( \text{A} \) as \( P_A^0 \). After the complete conversion of \( \text{A} \), the total pressure is 300 torr, thereby making the change in pressure due to reaction equal to \( 300-200=100 \, \text{torr} \) at 23 seconds similar: 
\( P_A = P_A^0 - x \) where \( x \) is the change in \( A \). Now, since every mole of \( \text{A} \) gives rise to 3 moles of products, at completion:\[(1-x) \rightarrow (x) \]The total pressure increase due to \( x \) of \( A \) decomposing to give \( 3x \) is proportional to formation. Simplify:\[P_{total}=P_A^0 + 2x = 200\]and at time \(t\) (23s):\[200=(P_A^0 -x) + 3x\]thus at equilibrium pushing,[\(3x\)] we re-derive: 
After substitution, equation implies using initial rate-integrated expression:\[k = \frac{2.303}{t} \log \left( \frac{P_\infty - P_0}{P_\infty - Pt} \right)\]
Given the pressures, it takes \(\log_{10}(2) = 0.301\), we substitute into the rate equation:
\[k = \frac{2.303}{23} \times \log \left( \frac{300-100}{300-200} \right)\]Evaluating with values:
\[k = \frac{2.303}{23} \times \log_{10}(2) \approx \frac{2.303}{23} \times 0.301\]
Approximating the result:
\[k \approx 0.301 \times 0.1 = 0.0303\]Thus, with the correct arithmetic:
\[k \approx 3 \times 10^{-2} \, \text{s}^{-1}\]

Answer 2

The reaction is: \[ \text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)} \]
Given:
\(P_{23} = P_0 + 2x = 200 \\  P_\infty = 3P_0 = 300 \\ P_0 = 100\)
The rate constant $K$ is calculated using:
\[ K = \frac{1}{t} \ln \frac{P_\infty - P_0}{P_\infty - P_t} \]
Substituting the values:
\[ K = \frac{2.3}{23} \log \frac{300 - 100}{300 - 200} \] \[ K = \frac{2.3 \times 0.301}{23} = 0.0301 = 3.01 \times 10^{-2} \, \text{s}^{-1} \]
The correct answer is (3).