Question

Calculate $\triangle G^ \ominus$and the equilibrium constant for the formation of $NO_{2}$ from NO and $O_{2}$ at 298K, $NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)$ where $\triangle G ^ \ominus (NO_{2}) = 52 .0 kJ/mol, \triangle G ^ \ominus(NO) = 87.0 kJ/mol, \triangle G ^ \ominus (O_{2})= 0 kJ/mol$

Answer 1

${ NO }_{ 2(g) }\quad +\quad \cfrac { 1 }{ 2 } { O }_{ 2(g) }\quad \rightarrow \quad { NO }_{ 2(g) }$

Given,

${ \Delta }_{ f }{ G }^{ \circ }\left( { NO }_{ 2 } \right) \quad =\quad 52.0\quad \sfrac { KJ }{ mol }$

${ \Delta }_{ f }{ G }^{ \circ }\left( { NO } \right) \quad =\quad 87.0\quad \sfrac { KJ }{ mol }$

${ \Delta }_{ f }{ G }^{ \circ }\left( { O }_{ 2 } \right) \quad =\quad 0\quad \sfrac { KJ }{ mol }$

We know that, ${ \Delta G }^{ \circ }\quad =\quad RT\quad \log { { K }_{ c } }$

${ \Delta G }^{ \circ }\quad =\quad 2.303\quad RT\quad \log { { K }_{ c } }$  

${ K }_{ c }\quad =\quad \frac { -35.0\quad \times \quad { 10 }^{ -3 } }{ -2.303\quad \times \quad 8.3114\quad \times \quad 298 } \quad$

By simplifying, we get Kc the equilibrium constant value $=\quad 1.36\quad \times \quad { 10 }^{ 6 }$