Question

In the reaction $2N_2{O}_{5\left(g\right)}⟶4N{O}_{2\left(g\right)}+O_{2(g{)}^2}$, initial pressure is $500$ atm and rate constant k is ${3.38\times 10}^{-5}se{c}^{-1}$. After $10$ minutes the final pressure of $N_2{O}_5$ is ?$(log1.02=0.0088)$

• A. $490$atm
• B. $250$ atm
• C. $480$ atm
• D. $420$ atm

Answer 1

The correct option is A $490$atm

The given reaction is

$2N_2{O}_5\stackrel{}{\to }4N{O}_2+O_2$

$P_0=500atm,k=3.38\times {10}^{-5}$

We know that

$k=\frac{2.303}{t}{\log }_{10}\frac{P_0}{P_t}$

$3.38\times {10}^{-5}{\sec }^{-1}=\frac{2.303}{10\times 60\sec }\log \frac{500}{P_t}$

$0.00880=\log \frac{500}{P_t}$

$P_t=490atm$

Hence, this is the required solution