Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="\mathrm{2O_{3}\left ( g \right )\rightleftharpoons 3O_{2}\left ( g \right )}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B2O_%7B3%7D%5Cleft%20%28%20g%20%5Cright%20%29

\mathrm{2O_{3}\left ( g \right )\rightleftharpoons 3O_{2}\left ( g \right )}

At \mathrm{300\, K}, ozone is fifty percent dissociated. The standard free energy change at this temperature and \mathrm{1} atm pressure is (-)_______ \mathrm{J\, mol^{-1}}. (Nearest integer)

[Given : \mathrm{ln\, 1.35= 0.3\; and\; R= 8.3\, J\, K^{-1}\, mol^{-1}}]

Option: 1

747


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

                  \mathrm{2 \mathrm{O}_{3} \rightleftharpoons 3 \mathrm{O}_{2} }

\mathrm{t=0 \quad\quad \mathrm{P}_{0} \quad \quad-}

\mathrm{t=t e q \quad \frac{P_{0}}{2} \quad \ \ \frac{3 P_{0}}{4}}

\mathrm{Given,\frac{P_{0}}{2}+\frac{3 P_{0}}{4}=1\; \Rightarrow \; P_{0}=\frac{4}{5}}

\mathrm{\therefore\; \mathrm{P_{O_{3}}}=\frac{2}{5} \; \; and \; \; \mathrm{P_{O_{2}}}=\frac{3}{5}}

\mathrm{\therefore \, K_{p}=\frac{P_{O_{2}}^{3}}{P_{O_{3}}^{2}}=\frac{(3 / 5)^{3}}{(2 / 5)^{2}}=\frac{27}{20}=1.35}

\mathrm{\therefore \, \, \Delta G=-R T \ln K =-8\cdot 3 \times 300 \times \ln (1\cdot 35)}
                                           \mathrm{=-8\cdot 3 \times 300 \times 0\cdot 3}
                                           \mathrm{=-747 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}}

Hence, the answer is 747

Posted by

Info Expert 30

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question