Question

The half life for the first order reaction ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}} \to {\rm{2N}}{{\rm{O}}_{\rm{2}}}\,{\rm{ + }}\,\dfrac{{\rm{1}}}{{\rm{2}}}{{\rm{O}}_{\rm{2}}} \ is \ 24 \ hrs \ at \ {\rm{3}}{{\rm{0}}^{\rm{0}}}{\rm{C}}$. Starting with \[{\rm{10}}\] g of \[{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}\], how many grams of \[{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}\]will remain after a period of \[96\]hrs.
A. \[{\rm{1}}{\rm{.25}}\,{\rm{g}}\]
B. \[0.63\,{\rm{g}}\]
C. \[1.77\,{\rm{g}}\]
D. \[0.5\,{\rm{g}}\]

Answer 1

Hint: Half change time or half- life period is the time during which the concentration of the reactant falls down to half of its initial value. This may also be defined as the time required for the half completion of the reaction. The half-life formula for a first order reaction can be written as $t_{\dfrac{1}{2}}= \dfrac{0.693}{k}$, where $t_{\dfrac{1}{2}}$ half-life and k = rate constant.

Formula used Integrated rate -law for first order reaction is k = $\dfrac{2.303}{t}.log(\dfrac{a}{a-x})$, where \[{\rm{k = }}\]rate constant
\[{\rm{t = }}\]time
\[{\rm{a = }}\]initial concentration of reactant at time \[{\rm{(t = 0)}}\]
\[{\rm{a}} - {\rm{x}}\,{\rm{ = }}\]concentration of the reactant after time \[{\rm{(t = t)}}\]

Complete Step by Step Solution:
A reaction is said to be of first order if the rate of the reaction depends upon one concentration term only.
For the reaction, \[{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}} \to {\rm{2N}}{{\rm{O}}_{\rm{2}}}\,{\rm{ + }}\,\dfrac{{\rm{1}}}{{\rm{2}}}{{\rm{O}}_{\rm{2}}}\]
Suppose the reaction is started with ‘\[{\rm{a}}\]’ moles per litre of the reactant \[{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}\]. After time \[{\rm{t}}\], suppose ‘\[{\rm{x}}\]’ moles per litre of it have decomposed. Therefore, the concentration of \[{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}\]after time \[{\rm{t}}\]\[{\rm{ = a - x}}\]moles per litre.

As per given data,
$t_{\dfrac{1}{2}} = 24 \ hrs$

Find the rate constant \[{\rm{(k)}}\] by using the half-life formula as:
$t_{\dfrac{1}{2}} = \dfrac{0.693}{k}$
$\Rightarrow k = \dfrac{0.693}{t_{\dfrac{1}{2}}}$
$\Rightarrow k = \dfrac{0.693}{24 \ hr}$
$\Rightarrow k = 0.0289 \ hr^{-1}$

Again,
a = 10
a - x = 10 - x
t = 96 hrs

Use integrated form of first order reaction to get the value of ‘\[{\rm{x}}\]’ as:
$k = \dfrac{2.303}{t}.log(\dfrac{a}{a-x})$
$\Rightarrow 0.0289 \ hr^{-1} = \dfrac{2.303}{96 \ hr}\times log(\dfrac{10}{10-x})$
$\Rightarrow 0.0289 \times 96 = 2.303\times log(\dfrac{10}{10-x})$
$\Rightarrow \dfrac{2.7744}{2.303} = log(\dfrac{10}{10-x})$
$\Rightarrow log(\dfrac{10}{10-x}) = 1.2$
$\Rightarrow \dfrac{10}{10-x} = antilog(1.2)$
$\Rightarrow \dfrac{10}{10-x} = 15.85$
$\Rightarrow \dfrac{10}{10-x} = 15.85 $
$\Rightarrow 10 = 158.5 - 15.85x $
$\Rightarrow x = \dfrac{148.5}{15.85} $
$\Rightarrow x = 9.37 g$
$\therefore a - x = (10 - 9.37)g = 0.63 \ g $
Therefore, option B is correct.

Note: The variation of concentration of reactants with time can be given by the integrated rate law equation. All reactions of the first order must obey the integrated rate law equation. The value of rate constant \[{\rm{(k)}}\]remains unchanged even if the concentration units are changed. The half -life for a first order reaction is found to be constant and independent of the initial concentration of the reactant.