Question

The following data were obtained during the first order thermal decomposition of $N_2{O}_5\left(g\right)$ at constant volume.

$2N_2{O}_5\left(g\right)\to 2N_2{O}_4\left(g\right)+O_2\left(g\right)$

Time (s)	0	100
Total pressure (atm)	0.5	0.512

Calculate the reaction rate when the total pressure is 0.52 atm.

• A. $Rate=2.26\times {10}^{-4}atm{s}^{-1}$
• B. $Rate=2.56\times {10}^{-4}atm{s}^{-1}$
• C. $Rate=2.37\times {10}^{-4}atm{s}^{-1}$
• D. $Noneofthese$

Answer 1

The correct option is A $Rate=2.26\times {10}^{-4}atm{s}^{-1}$

a. $2N_2{O}_5\left(g\right)\to 2N_2{O}_4\left(g\right)+O_2\left(g\right)$

(OR)

$N_2{O}_5\to N_2{O}_4+\frac{1}{2}{O}_2$

Initial $0.500$

Final $(0.5-x)x\frac{x}{2}$

$P_t=0.5-x+x+\frac{x}{2}=0.5+\frac{x}{2}$.

$0.5+\frac{x}{2}=0.512$

$x=0.024$

Now,

$k=\frac{2.303}{t}log\frac{a}{0.5-x}$

$k=\frac{2.303}{100}log\frac{0.5}{0.5-0.024}$

$k=\frac{2.303}{100}log\frac{0.5}{0.476}$

$k=\frac{2.303}{100}\times 0.0214=4.92\times {10}^{-4}{s}^{-1}$

b. $P_t=0.52atm$

$0.5+\frac{x}{2}=0.52\Rightarrow x=0.04$

$a-x=0.5-0.04=0.46$ atm

Rate $=k{P}_{N_2{O}_5}$

Rate $=4.92\times {10}^{-4}{s}^{-1}\times 0.46=2.26\times {10}^{-4}atm{s}^{-1}$

Hence, the correct option is $\text{A}$