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  • The standard free energy change <img alt="\left ( \Delta \text{G}^{\text{o}} \right )" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%20%28%20%5CDelta%20%5Ctext%7BG%7D%5E%7B%5Ctext%7B

The standard free energy change \left ( \Delta \text{G}^{\text{o}} \right ) for 50% dissociation of \mathrm{N_{2}O_{4}} into \mathrm{NO_{2}} at \mathrm{27^{\text{o}}C} and 1 atm pressure is -x \mathrm{~J} \mathrm{~mol}^{-1}. The value of x is _________.(Nearest Integer)

\text { [Given }: \mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 1.33=0.1239 \ln 10=2.3 \text { ] }

 

Option: 1

710


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} &\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2} \\ &\mathrm{P}_{\mathrm{o}} \\ &\frac{\mathrm{P}_{o}}{2}\; \; \; \; \; \; \; \; \mathrm{P_{o}} \end{aligned}

Given,

\begin{aligned} &\quad P_{0}+\frac{P_{0}}{2}=1\\ &\Rightarrow \frac{3 P_{0}}{2}=1 \Rightarrow P_{0}=\frac{2}{3}\\ &\therefore P_{\mathrm{NO}_{2}}=\frac{2}{3} ; \quad P_{\mathrm{N}_{2} \mathrm{O}_{4}}=\frac{1}{3} \end{aligned}

K_{p}=\frac{P_{\mathrm{NO}_{2}}{ }^{2}}{P_{\mathrm{N}_{2} \mathrm{O}_{4}}}=\frac{(2 / 3)^{2}}{(1 / 3)}=\frac{4}{3}

\begin{aligned} \therefore \Delta G^{0}=-R T \ln K &=-8.31 \times 300 \times \ln \left(\frac{4}{3}\right) \\ &=-8.31 \times 300 \times 2.3 \times 0.1239 \\ &=-710.4 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned}

Hence, the answer is 710

Posted by

vinayak

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