Question

Decomposition of both $A_2\left(g\right)$ and $B_3\left(g\right)$ follows $1^{st}$ order kinetics as:
$A_2\left(g\right)\stackrel{k_1}{\to }2A\left(g\right)k_1\left(h{r}^{-1}\right)={10}^2{e}^{-\frac{14000}{RT}}$
$B_3\left(g\right)\stackrel{k_2}{\to }3B\left(g\right)k_2\left(h{r}^{-1}\right)={10}^3{e}^{-\frac{20000}{RT}}$
If $1mol$ of $A_2\left(g\right)$ and $B_3\left(g\right)$ are taken in a $10L$ evacuated flask and heated to some temperature so that they start decomposing at the same rate, determine total pressure (in $atm$) in the flask after $1.0hr$.

Answer 1

Since, $A_2\left(g\right)$ and $B_3\left(g\right)$ start decompositing at the same rate.
$\therefore r_{A_2\left(g\right)}=r_{B_3\left(g\right)}$
$k_1\left[A_2\right]=k_2\left[B_3\right]k_1\frac{1}{10}=k_2\frac{1}{10}\therefore k_1=k_2{10}^2{e}^{-\frac{14000}{RT}}={10}^3{e}^{-\frac{20000}{RT}}$

$\therefore T=313.42K$
$k_1={10}^2{e}^{-\frac{14000}{RT}}{k}_1={10}^2{e}^{-\frac{14000}{8.314\times 313.42}}\therefore k_1=0.464h{r}^{-1}$
$\because k_1=k_2\therefore k_2=0.464h{r}^{-1}$

$\Rightarrow k_{1}t=\text{ln}\frac{n_0\left(A_2\right)}{n_0\left(A_2\right)-n_1}\Rightarrow n_1=0.37\Rightarrow k_{2}t=\text{ln}\frac{n_0\left(B_3\right)}{n_0\left(B_3\right)-n_1}\Rightarrow n_2=0.37$

$A_2\left(g\right)\to 2A\left(g\right)t=010t=1.0hr1-n_{1}2n_1$

$\text{Number of moles of gases after}1.0hr=1-n_1+2n_1=1+n_1=1+0.37=1.37$

$B_3\left(g\right)\to 3B\left(g\right)t=010t=1.0hr1-n_{2}3n_2$

$\text{Number of moles of gases after}1.0hr=1-n_2+3n_2=1+2n_2=1+2\left(0.37\right)=1.74$

$\text{Total number of moles of gases after}1.0hr\left(n_T\right)=1.37+1.74=3.11$

$P_T=\frac{n_{T}RT}{V}=\frac{3.11\times 0.0821\times 313.42}{10}=8atm$