Question

A ${N_2}{O_5}$decomposes to $N{O_2}$ and ${O_2}$ and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from $50mmHg$to $87.5mmHg$. The pressure of the gaseous mixture after 100 minutes at constant temperature will be __________
A.$116.25mmHg$
B.$106.25mmHg$
C.$136.25mmHg$
D.$175.0mmHg$

Answer 1

The first order kinetic reaction is a reaction in which the rate of reaction depends on the concentration of only one reactant and is proportional or linearly related to the amount of reactant.
Rate = $- \dfrac{{d\left[ A \right]}}{{dt}} = k{\left[ A \right]^1} = k\left[ A \right]$

Complete step by step solution: ${N_2}{O_5} \to 2N{O_2} + \dfrac{1}{2}{O_2}$
$R = k\left[ {{N_2}{O_5}} \right]$

1.	${P_0}$	0(initial)	0(final)	At t=0minute
2.	${P_0} - p$	$2p$	$\dfrac{1}{2}p$	At t=50minute
3.	${P_0} - {p^1}$	$2{p^1}$	$\dfrac{1}{2}{p^1}$	At t=100minute

Given, ${P_0} = 50mmHg$
So, ${P_{50\min }} = {P_0} + \dfrac{3}{2}p = 87.5$
Solving for $p$

$$p = \dfrac{2}{3}(87.5 - 50) \\\ p = \dfrac{{37.5}}{3} \times 2 = 25mmHg \\\$$

First order equation,
$t = \dfrac{{2.303}}{k}\log \dfrac{{{{\left( {{N_2}{O_5}} \right)}_{t = 0}}}}{{{{\left( {{N_2}{O_5}_{}} \right)}_t}}}$ - (1)
At t=50 minutes
Equation (1) becomes$t = \dfrac{{2.303}}{k}\log 2$
So, $k = \dfrac{{2.303}}{{50}} \times 0.3010$
At t=100 minutes
Equation (1) becomes
$t = \dfrac{{2.303 \times 50}}{{2.303 \times 0.3010}}\log \dfrac{{{{\left( {{N_2}{O_5}} \right)}_{t = 0}}}}{{{{\left( {{N_2}{O_5}} \right)}_{t = 100}}}}$

$$100 = \dfrac{{2.303 \times 50}}{{2.303 \times 0.3010}}\log \dfrac{{{{\left( {{N_2}{O_5}} \right)}_{t = 0}}}}{{{{\left( {{N_2}{O_5}} \right)}_{t = 100}}}} \\\ 2 \times 0.3010 = {\log _{10}}\dfrac{{50}}{x} \\\ so,x = \dfrac{{50}}{4} \\\$$

Therefore, ${p^1} = 50 - 12.5 = 37.5mmHg$
So, total pressure= ${P_0} + \dfrac{3}{2} \times 37.5$
= $106.25mmHg$
Thus, the correct answer for the given question is option B.

Note: The second order kinetic reaction is a complicated and difficult reaction because it involves measuring the two reactants simultaneously and has other complications due to the rate of reaction or any property of the reactants. Thus, to avoid the expensive and complicated reactions, second order reactions can be treated as pseudo first order reactions.
$- \dfrac{{d\left[ A \right]}}{{dt}} = k\left[ A \right]\left[ B \right]$