Question

The half-life decomposition of ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}$is a first order reaction represented by ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}} \to {{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}} + 1/2{{\rm{O}}_{\rm{2}}}$
After 15 min the volume of ${{\rm{O}}_{\rm{2}}}$produced in $9\;{\rm{mL}}$ and at the end of the reaction $35\;{\rm{mL}}$. The rate constant is equal to:
A. $\dfrac{1}{{15}}\ln \dfrac{{35}}{{26}}$
B. $\dfrac{1}{{15}}\log \dfrac{{44}}{{26}}$
C. $\dfrac{1}{{15}}\log \dfrac{{35}}{{36}}$
D. None of the above

Answer 1

For the reaction, ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}} \to {{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}} + 1/2{{\rm{O}}_{\rm{2}}}$, first we have to write the ICE table from the given information, such as, amount of ${{\rm{O}}_{\rm{2}}}$produced after 15 minutes and at the end of the reaction.

Complete step-by-step answer:
We take ${A_0}$ as the initial concentration of ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}$and x as the change of concentration of ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}$. Now, we construct the Initial change Equilibrium (ICE) table.

For the reaction, ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}} \to {{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}} + 1/2{{\rm{O}}_{\rm{2}}}$

| ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}$| ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}$| $1/2{{\rm{O}}_{\rm{2}}}$
At t=0| ${A_0}$| 0| 0
at t=15| ${A_0} - x$| x| x/2 (${\rm{9 mL}}$)
At t=$\infty$| 0| ${A_0}$| $\dfrac{{{A_0}}}{2}$(${\rm{35 mL}}$)

Given that, after 15 minutes, the amount of oxygen is ${\rm{9 mL}}$. Now, we calculate the value of x from the ICE table.

$\begin{array}{l}\dfrac{x}{2} = 9\\\x = 18\end{array}$

So, the value of x is 18.

At the end of the reaction, the amount of oxygen is ${\rm{35 mL}}$. So, from the fourth row of the ICE table we calculate the value of ${A_0}$.

$\begin{array}{l}\dfrac{{{A_0}}}{2} = 35\\\\{A_0} = 70\end{array}$

Now, we use the rate constant equation.

$\begin{array}{l}k = \dfrac{1}{t}\ln \dfrac{{\left[ {{A_0}} \right]}}{{\left[ {{A_t}} \right]}}\\\k = \dfrac{1}{t}\ln \dfrac{{{A_0}}}{{{A_0} - x}}\end{array}$ …… (1)

Here, k is rate constant, t is time, $\left[ {{A_0}} \right]$is initial concentration and $\left[ {{A_t}} \right]$is final concentration.

Now, we put the Value ${A_0}$and x in the equation (1). We also put t=15 in the equation.

$\begin{array}{c}k = \dfrac{1}{{15}}\ln \dfrac{{70}}{{70 - 18}}\\\ = \dfrac{1}{{15}}\ln \dfrac{{70}}{{52}}\\\ = \dfrac{1}{{15}}\ln \dfrac{{35}}{{26}}\end{array}$

So, the rate constant for the reaction is $K = \dfrac{1}{t}\ln \dfrac{{35}}{{26}}$. Hence, the correct option is A.

Note: In the ICE table, students might think that after 15 minutes, amount of ${{\rm{O}}_{\rm{2}}}$formed is x, but actually it is x/2 as $1/2{{\rm{O}}_{\rm{2}}}$is present in the chemical equation. If these values are taken correctly, equilibrium constant can be calculated correctly.