Question

The decomposition of ${{N}_{2}}{{O}_{5}}$ according to the equation $2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g)$, is a first order reaction. After 30 minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be 284.5mm of$Hg$and on complete decomposition, the total pressure is 584.5mm of $Hg$. The rate constant of a reaction is:
A)$5.206\times {{10}^{-3}}{{\min }^{-1}}$
B) $4.261\times {{10}^{-3}}{{\min }^{-1}}$
C) $3.316\times {{10}^{-4}}{{\min }^{-1}}$
D) None of these

Answer 1

Hint The answer to this question is based on the rate constant of a reaction formula for the first order reaction according to the kinetics which is given by, $k=\dfrac{2.303}{t}\log \dfrac{{{P}_{0}}}{{{P}_{f}}}$ and this leads to the required answer.

Complete answer:
In the lower classes of physical chemistry, we have come across the chapter which deals with the kinetics of the reactions in solution where we have studied the concepts of finding the order of the reaction rate constant and also several other required entities.
Let us now focus on the rate constant which is to be found based on the pressure data which is given.
Now, the decomposition reaction given is,
$2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g)$

t = 0	${{P}_{1}}$	0	0
t = 30	${{P}_{1}}-2x$	+ 4x	+ x
t = $\infty $	${{P}_{1}}$-${{P}_{1}}$= 0	2${{P}_{1}}$	${}^{{{P}_{1}}}/{}_{2}$

The reaction at infinity is nothing but the completion of the reaction. Thus, the product pressure reduces to half value.
Now, according to the data given,
Total pressure at t = 30 min we have,
${{P}_{1}}-2x+4x+x=284.5mm $of $Hg$
\[\Rightarrow {{P}_{1}}+3x=284.5mm\]of $Hg$ …………….(1)
When the reaction completes, the total pressure according to the data given is,
\[2{{P}_{1}}+{}^{{{P}_{1}}}/{}_{2}=584.5mm\]of $Hg$
\[\Rightarrow \dfrac{5{{P}_{1}}}{2}=584.5mm\]of $Hg$
\[\Rightarrow {{P}_{1}}=\dfrac{2\times 584.5}{5}=233.8mm\]of $Hg$
Now, we have to find the value of x.
From equation (1),
\[{{P}_{1}}+3x=284.5mm\] of $Hg$
Now, substituting the value of ${{P}_{1}}$ in this equation number (1), we have
\[233.8mm(Hg)+3x=284.5mm(Hg)\]
\[\Rightarrow 3x=50.7mm\]of $Hg$
Thus, by simplification of the above equation, we have
\[x=16.9mm\] of $Hg$
Now, the rate constant for the first order reaction in terms of initial and final pressure is given by,
$k=\dfrac{2.303}{t}\log \dfrac{{{P}_{0}}}{{{P}_{f}}}$ ……………(2)
Here, we know that${{P}_{0}}=233.8mm $of $Hg$
Now the final pressure thus, will be
\[{{P}_{f}}={{P}_{1}}-2x\]
\[\Rightarrow 233.8mm(Hg)-2\times 16.9mm(Hg)=200mm\] of$Hg$
Now, substituting this in above equation (2),
$k=\dfrac{2.303}{30}\log \dfrac{233.8mm(Hg)}{200mm(Hg)}$
$k=0.00521{{\min }^{-1}}=5.21\times {{10}^{-3}}{{\min }^{-1}}$

Thus, the correct answer is option A) $5.206\times {{10}^{-3}}{{\min }^{-1}}$

Note: Note that the rate constant of the first order or the ${{n}^{th}}$ order reaction can be written in other forms also. In terms of the percentage decomposition data the formula will be $k=\dfrac{2.303}{t}\log \dfrac{a}{a-x}$ and by the rate constant formula, you can find the half life of the first order reaction and also the ${{n}^{th}}$ order reaction