question_answer 1)

Calculate (a)\[\Delta {{G}^{{}^\circ }}\]and (b) the equilibrium constant for the formation of \[N{{O}_{2}}\] from NO and \[{{O}_{2}}\]at \[298\,K\] \[NO(g)+\frac{1}{2}{{O}_{2}}(g)\rightleftharpoons N{{O}_{2}}(g)\]where\[{{\Delta }_{f}}{{G}^{{}^\circ }}(N{{O}_{2}})=52kJ/mol;\,\]\[{{\Delta }_{f}}{{G}^{{}^\circ }}(NO)\] \[=87kJ/mol;\]\[{{\Delta }_{f}}{{G}^{{}^\circ }}({{O}_{2}})=0kJ/mol\]

Answer:

(a)\[\Delta G_{\operatorname{Re}action}^{{}^\circ }=\Sigma {{\Delta }_{f}}G_{\text{Products}}^{\text{ }\!\!{}^\circ\!\!\text{ }}-\Sigma {{\Delta }_{f}}G_{\text{Reactants}}^{\text{ }\!\!{}^\circ\!\!\text{ }}\] \[={{\Delta }_{f}}{{G}^{{}^\circ }}N{{O}_{2}}-[{{\Delta }_{f}}{{G}^{{}^\circ }}NO+\frac{1}{2}{{\Delta }_{f}}{{G}^{{}^\circ }}{{O}_{2}}]\] \[=52-[87+\frac{1}{2}\times 0]=-35kJ\] (b)\[{{K}^{{}^\circ }}=Anti\log \left( \frac{-\Delta {{G}^{{}^\circ }}}{2.303RT} \right)\] \[=Anti\log \left( \frac{35000}{2.303\times 8.314\times 298} \right)\] \[=1.36\times {{10}^{6}}\]