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Question: The decomposition of \({{N}_{2}}{{O}_{5}}\) according to the equation \(2{{N}_{2}}{{O}_{5}}(g)\to 4N...

The decomposition of N2O5{{N}_{2}}{{O}_{5}} according to the equation 2N2O5(g)4NO2(g)+O2(g)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 ofHgHgand on complete decomposition, the total pressure is 584.5mm of HgHg. The rate constant of a reaction is:
A)5.206×103min15.206\times {{10}^{-3}}{{\min }^{-1}}
B) 4.261×103min14.261\times {{10}^{-3}}{{\min }^{-1}}
C) 3.316×104min13.316\times {{10}^{-4}}{{\min }^{-1}}
D) None of these

Explanation

Solution

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=2.303tlogP0Pfk=\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,
2N2O5(g)4NO2(g)+O2(g)2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g)

t = 0P1{{P}_{1}}00
t = 30P12x{{P}_{1}}-2x+ 4x+ x
t = \infty P1{{P}_{1}}-P1{{P}_{1}}= 02P1{{P}_{1}}P1/2{}^{{{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,
P12x+4x+x=284.5mm{{P}_{1}}-2x+4x+x=284.5mm of HgHg
P1+3x=284.5mm\Rightarrow {{P}_{1}}+3x=284.5mmof HgHg …………….(1)
When the reaction completes, the total pressure according to the data given is,
2P1+P1/2=584.5mm2{{P}_{1}}+{}^{{{P}_{1}}}/{}_{2}=584.5mmof HgHg
5P12=584.5mm\Rightarrow \dfrac{5{{P}_{1}}}{2}=584.5mmof HgHg
P1=2×584.55=233.8mm\Rightarrow {{P}_{1}}=\dfrac{2\times 584.5}{5}=233.8mmof HgHg
Now, we have to find the value of x.
From equation (1),
P1+3x=284.5mm{{P}_{1}}+3x=284.5mm of HgHg
Now, substituting the value of P1{{P}_{1}} in this equation number (1), we have
233.8mm(Hg)+3x=284.5mm(Hg)233.8mm(Hg)+3x=284.5mm(Hg)
3x=50.7mm\Rightarrow 3x=50.7mmof HgHg
Thus, by simplification of the above equation, we have
x=16.9mmx=16.9mm of HgHg
Now, the rate constant for the first order reaction in terms of initial and final pressure is given by,
k=2.303tlogP0Pfk=\dfrac{2.303}{t}\log \dfrac{{{P}_{0}}}{{{P}_{f}}} ……………(2)
Here, we know thatP0=233.8mm{{P}_{0}}=233.8mm of HgHg
Now the final pressure thus, will be
Pf=P12x{{P}_{f}}={{P}_{1}}-2x
233.8mm(Hg)2×16.9mm(Hg)=200mm\Rightarrow 233.8mm(Hg)-2\times 16.9mm(Hg)=200mm ofHgHg
Now, substituting this in above equation (2),
k=2.30330log233.8mm(Hg)200mm(Hg)k=\dfrac{2.303}{30}\log \dfrac{233.8mm(Hg)}{200mm(Hg)}
k=0.00521min1=5.21×103min1k=0.00521{{\min }^{-1}}=5.21\times {{10}^{-3}}{{\min }^{-1}}

Thus, the correct answer is option A) 5.206×103min15.206\times {{10}^{-3}}{{\min }^{-1}}

Note: Note that the rate constant of the first order or the nth{{n}^{th}} order reaction can be written in other forms also. In terms of the percentage decomposition data the formula will be k=2.303tlogaaxk=\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 nth{{n}^{th}} order reaction