Discussion Forum



1)

 The following reaction is performed at 298K $2NO(g)+O_{2}(g)\rightleftharpoons 2NO_{2}(g)$. The standard free energy of formation of NO (g)  is 86.6kJ/mol at 298K. What is the standard free energy of formation of NO2 (g)  at 298K? ($K_{p}=1.6\times 10^{12}$)


A) R(298) in ( $1.6\times 10^{12}$)-86600

B) 86600+R(298)ln ( $1.6\times 10^{12}$)

C) 86600- $\frac{ln(1.6\times 10^{12})}{R(298)}$

D) $0.5 [2\times 86600- R(298)ln (1.6\times 10^{12})]$

Answer:

Option D

Explanation:

 For the given reaction,

     $2NO(g)+O_{2}(g)\rightleftharpoons 2NO_{2}(g)$

 Given,   $\triangle G_{f}^{0} (NO)=86.6 kJ/mol$

    $\triangle G_{f}^{0} (NO_{2})=?$

              $K_{p}=1.6\times 10^{12}$

Now , we have

   $\triangle G_{f}^{0}=2\triangle  G_{f(NO_{2})}^{0}-[2\triangle G_{f(NO)}^{0}+ \triangle G_{f(O_{2})}^{0}]$

                  $-RT ln K_{p}=2\triangle  G_{f(NO_{2})}^{0} -(2\times 86600+0)$

   $\triangle  G_{f(NO_{2})}^{0}=\frac{1}{2}$

                                                            $[2\times 86600-R\times298 ln (1.6\times 10^{12}]$

   $\triangle  G_{f(NO_{2})}^{0}=0.5$

                                                   $[2\times 86600-R\times(298) ln (1.6\times 10^{12})]$