Question

The half-life of decomposition of $N_2{O}_5$ is a first order reaction represented by:

$N_2{O}_5\to N_2{O}_4+1/2O_2$

After 15 minutes, the volume of $O_2$ produced is 9 $mL$ and at the end of the reaction is 35 $mL$. The rate constant is equal to:

• A. $\frac{1}{15}lo{g}_e\frac{35}{26}$
• B. $\frac{1}{15}lo{g}_e\frac{44}{26}$
• C. $\frac{1}{15}lo{g}_e\frac{35}{36}$
• D. None of the above

Answer 1

Answer: (A)

$\frac{1}{15}lo{g}_e\frac{35}{26}$

For a first order reaction,

$KT=ln(a/a-x)$

So, for the following reaction:

$N_2{O}_5\to N_2{O}_4+1/2O_2$

$K\times 15=ln\left(\begin{array}{c}\frac{35-0}{35-9}\end{array}\right)$

$N_2{O}_5\to N_2{O}_4+1/2O_2$

$K=\frac{1}{15}ln\left(\begin{array}{c}\frac{35}{26}\end{array}\right)$