Question

The reaction $A\left(g\right)+2B\left(g\right)\to C\left(g\right)+D\left(g\right)$ is an elementary process. In an experiment, the initial partial pressure of $A$ and $B$ are $p_A=0.60$ and $p_B=0.80atm$. When $p_C=0.2atm$ and $p_D=0.8atm$ , the ratio of the rate of reaction to the initial rate is:

• A. $\frac{1}{48}$
• B. $\frac{5}{12}$
• C. $\frac{9}{16}$
• D. $\frac{1}{6}$

Answer 1

Answer: (D)

$\frac{1}{6}$

The given reaction is:-

$A\left(g\right)+2B\left(g\right)⟶C\left(g\right)+D\left(g\right)$

Since, it is an elementary process, so the rate law is given by:-

Rate=$K\left[A\right][B{]}^2$; $K$ is rate constant

Given, Initial partial pressure of $A$=$0.6$ atm

Initial partial pressure of $B$=$0.8$ atm

Now, $A\left(g\right)+2B\left(g\right)⟶C\left(g\right)+D\left(g\right)$

$t=0$ $0.6$ $0.8$ $0$ $0$

$t=t$ $0.6-P$ $0.8-2P$ $P$ $P$

Case I:-

Now, given $Pc=0.2$ atm $=P$

So, $P_A=0.4$ atm , $P_B=0.4$ atm

$\therefore$ Initial rate=$R_i=K\left[A\right][B{]}^2$

$R_i=K\times 0.6\times 0.8\times 0.8$

Final rate=$R_f=K\times 0.4\times 0.4\times 0.4$

$\therefore \frac{R_f}{R_i}=\frac{0.4\times 0.4\times 0.4}{0.6\times 0.8\times 0.8}=\frac{1}{6}$