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  • The following data were obtained during the first order thermal decomposition of <img alt="\mathrm{N}_2 \mathrm{O}_5(\mathrm{~g})" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BN

The following data were obtained during the first order thermal decomposition of \mathrm{N}_2 \mathrm{O}_5(\mathrm{~g}) at constant volume:

2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 2 \mathrm{~N}_2 \mathrm{O}_4(g)+\mathrm{O}_2(g)

S.No.     Time/s         Total Pressure/(atm)

1.             0                    0.5

2.            100                 0.512

Calculate the rate constant.

Option: 1

0.000598 s-1


Option: 2

0.000698 s-1


Option: 3

0.000598 s-1


Option: 4

0.000491 s-1


Answers (1)

best_answer

:Suppose the pressure of \mathrm{N}_2 \mathrm{O}_5(\mathrm{~g}) decreases by2 \mathrm{x\: atm}. As per the balanced chemical equation, two moles of\mathrm{N}_2 \mathrm{O}_5 decompose to give two moles of \mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) and one mole of \mathrm{O}_2(\mathrm{~g}). Hence, the pressure of \mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) increases by 2 \mathrm{x\: atm} and the pressure of \mathrm{O}_2(\mathrm{~g}) increases by x \mathrm{~atm}.
The given equation represents the decomposition of dinitrogen pentoxide gas into nitrogen dioxide and oxygen gases:

2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 2 \mathrm{~N}_2 \mathrm{O}_4(g)+\mathrm{O}_2(g)

At the initial time t=0, the pressures of \mathrm{N}_2 \mathrm{O}_5, \mathrm{~N}_2 \mathrm{O}_4, and \mathrm{O}_2(\mathrm{~g}) are 0.5 \mathrm{~atm}, 0 \mathrm{~atm}, and 0 \mathrm{~atm} respectively. Let the pressure of \mathrm{\mathrm{N}_2 \mathrm{O}_5\: decrease \: by\: 2 \mathrm{x} \: atm \: at \: time \ \mathrm{t}}

After the decomposition, two moles of \mathrm{N}_2 \mathrm{O}_5  produce two moles of \mathrm{N}_2 \mathrm{O}_4 and one mole of \mathrm{O}_2. Therefore, the pressure of \mathrm{N}_2 \mathrm{O}_4 increases by 2 \mathrm{x}$ atm and that of $\mathrm{O}_2(\mathrm{~g}) increases by \mathrm{x} atm.

Thus, at time t, the pressure of \mathrm{N}_2 \mathrm{O}_5 will be (0.5-2 \mathrm{x}) atm, the pressure of \mathrm{N}_2 \mathrm{O}_4 will be 2 x atm and the pressure of \mathrm{O}_2(\mathrm{~g}) will be x atm.
Total pressure = pressure or all gasse =p

\\(0.5-2 x)+2 x+x=0.5+x=P\\ x=P-0.5

\\\ \mathrm{Pressure \: of\; \mathrm{N}_2 \mathrm{O}_5=0.5-2 x}\\ =0.5-2(P-0.5)=1.5-2 P\\ At \ \mathrm{t}=100 \mathrm{~s} ; \mathrm{P}=0.512 \mathrm{~atm}

\\\mathrm{Pressure\: of \: \mathrm{N}_2 \mathrm{O}_5=1.5-2 \times 0.512=0.476 \mathrm{~atm}} \\ \\\mathrm{Using\:\; the\: \; formula\: \; of \; \mathrm{k}\; is: K} \\\\=\frac{2.303}{t} \log \frac{\text { initial pressure }}{\text { final pressure }}=\frac{2.303}{t} \log \frac{0.5}{0.476}=0.000491 \mathrm{sec}^{-1}

Posted by

HARSH KANKARIA

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