Question

For a first order gas phase reaction-
$A_{\left(g\right)}\to 2B_{\left(g\right)}+C_{\left(g\right)}$
P₀ be initial pressure of A and P_t the total pressure at time 't'. Integrated rate equation is

• A. $\frac{2.303}{t}log\left(\frac{P_0}{P_0-P_t^{÷}}\right)$
• B. $\frac{2.303}{t}log\left(\frac{2P_0}{3P_0-P_t^{÷}}\right)$
• C. $\frac{2.303}{t}log\left(\frac{P_0}{2P_0-P_t^{÷}}\right)$
• D. $\frac{2.303}{t}log\left(\frac{2P_0}{2P_0-P_t^{÷}}\right)$

Answer 1

Answer: (B)

Initial : $\begin{array}{ccccc}{A}_{\left(g\right)}& \to & 2B_{\left(g\right)}& +& C_{\left(g\right)}\\ P_0& & 0& & 0\\ P_0-P& & 2P& & P\end{array}$

Total pressure at time $(t)=P_0-P+2P+P=P_t$

$\Rightarrow P_t=P_0+2P$

$P_t=P_0+2P$

$\Rightarrow P=\frac{P_t-P_0}{2}$

$k=\frac{2.303}{t}log\left[\frac{P_0}{P_0-P}\right]$

$=\frac{2.303}{t}log\left[\frac{P_0}{P_0-\left(\frac{P_t-P_0}{2}÷\right)}\right]$

$=\frac{2.303}{t}log\left(\frac{2P_0}{2P_0-P_t+P_0}\right)$

$=\frac{2.303}{t}log\left(\frac{2P_0}{3P_0-P_t}\right)$