Question

For a first order gas phase reaction- $A_{\left(g\right)} \to 2B_{\left(g\right)} + C_{\left(g\right)}$ P be initial pressure of A and P the total pressure at time 't'. Integrated rate equation is

• A. $\frac{2.303}{t}log \left(\frac{P_{0}}{P_{0}-P ^{\div}_{t}}\right)$
• B. $\frac{2.303}{t}log \left(\frac{2P_{0}}{3P_{0}-P^{\div}_{t}}\right)$
• C. $\frac{2.303}{t}log \left(\frac{P_{0}}{2P_{0}-P^{\div}_{t}}\right)$
• D. $\frac{2.303}{t}log \left(\frac{2P_{0}}{2P_{0}-P^{\div}_{t}}\right)$

Answer 1

Answer: (B)

$\frac{2.303}{t}log \left(\frac{2P_{0}}{3P_{0}-P^{\div}_{t}}\right)$

Answer 2

Initial : $\begin{matrix}A_{\left(g\right)}&\to&2B_{\left(g\right)}&+&C_{\left(g\right)}\\\ P_{0}&&0&&0\\\ P_{0}-P&&2P&&P\end{matrix}$ Total pressure at time $(t) = P_{0} - P + 2P + P = P_t$ $\Rightarrow P_{t} = P_{0} + 2P$ $P_{t} = P_{0} + 2P$ $\Rightarrow P=\frac{P_{t}-P_{0}}{2}$ $k=\frac{2.303}{t}log\left[\frac{P_{0}}{P_{0}-P}\right]$ $=\frac{2.303}{t}log\left[\frac{P_{0}}{P_{0}-\left(\frac{P_{t}-P_{0}}{2}\div\right)}\right]$ $=\frac{2.303}{t}log\left(\frac{2P_{0}}{2P_{0}-P_{t}+P_{0}}\right)$ $=\frac{2.303}{t}log\left(\frac{2P_{0}}{3P_{0}-P_{t}}\right)$