Question

Decomposition of both $A_2\left(g\right)$ and $B_3\left(g\right)$ follows first order kinetics as:
$A_2\left(g\right)\stackrel{k_1}{\to }2A\left(g\right);k_1\left(perh{r}^{-1}\right)=e^{\frac{14000\left(/J\right)}{RT}+5}{B}_3\left(g\right)\stackrel{k_2}{\to }3B\left(g\right);k_2\left(perh{r}^{-1}\right)=e^{\frac{20000\left(/J\right)}{RT}+10}$
Where $k_1$ and $k_2$ are the rate constants with respect to disappearance of $A_2$ and $B_3$, respectively.
One mole of each of $A_2\left(g\right)$ and $B_3\left(g\right)$ are taken in a $100\text{L}$ evacuated flask and at some temperature at which they start decomposing at the same rate. Select incorrect statement(s):
(Here $A_2$ and $B_3$ gases are nor reacting ideal gases)

• A. The temperature at which the rections are performed is $\frac{1200}{R}$
• B. At any instant, $\left(\frac{P_{A_2}}{P_{B_3}}\right)$ will be constant and equal to $1$ where $P_{A_2}=$ Partial pressure of $A_2$ and $P_{B_3}=$ Partial pressure of $B_3$
• C. At some instant, the total pressure of all the gases may be less than $0.2$ atm
• D. At any instant, $\left(\frac{P_A}{P_B}\right)$ will be constant and equal to 1 where $P_A=$ Partial pressure of $A$ and $P_B=$ Partial pressure of $B$

Answer 1

Answer: (C), (D)

At any instant, $\left(\frac{P_A}{P_B}\right)$ will be constant and equal to 1 where $P_A=$ Partial pressure of $A$ and $P_B=$ Partial pressure of $B$

(A) Given
$A_2\left(g\right)\to 2A\left(g\right);k_1=e^{\frac{14000\left(/J\right)}{RT}+5}{B}_3\left(g\right)\to 3B\left(g\right);k_2=e^{\frac{20000\left(/J\right)}{RT}+10}\text{So}{r}_2=k_2\left[B_3\right]\text{Rate}\left(r_1\right)=k_1\left[A_2\right]\text{Given the rates of the reactions are same}\therefore r_1=r_2\Rightarrow k_1\left[A_2\right]=k_2\left[B_3\right]\text{Again since we have taken 1 mol of eact of the reactant, hence}\therefore k_1=k_2$
$\Rightarrow e^{-\frac{14000}{RT}+5}=e^{-\frac{20000}{RT}+10}\text{Taking log on both sides}\frac{-14000}{RT}+5=-\frac{20000}{RT}+10\Rightarrow -14000+5RT=20000+10RT\Rightarrow 6000=5RT\Rightarrow T=\frac{1200}{R}$

(B) Initial moles of $A_2$ and $B_3$ are same and rate of disappearance of $A_2$ and $B_3$ is also same thus at any instant, $\left(\frac{P_{A_2}}{P_{B_3}}\right)$ will be constant and equal to $1$

(C) From ideal gas equation $PV=nRT$
Here we are considering total $2$ moles of gases, in $100\text{L}$ evacuated flask, so
$P=\frac{nRT}{V}=2\times \frac{0.0821}{100}\times \frac{1200}{8.3}\Rightarrow P=0.237\text{atm}$
Since the intial pressure is $0.237\text{atm}$ , hence after decomposition the pressure will increase only. Thus at any instant the total pressure of all the gases cannot be $0.2\text{atm}$

(D) Since the decomposed quantity is different for both $A$ and $B$ and the decomposition rates are same thus $\left(\frac{P_A}{P_B}\right)$ will not be constant and equal to 1 at any point of time.