Question

What is the equilibrium constant K for the following reaction at 400 K? $2NOCl(g) \rightleftharpoons 2NO(g)+Cl_{2}(g)$, given: ΔH=77.2 kJ/mol and ΔS=122 J $K^{-1}$ $mol^{-1}$ at 400K.

Answer 1

We know that, ${ \Delta G }^{ \circ  }\quad =\quad { \Delta H }^{ \circ}\quad -\quad T{ \Delta S }^{ \circ}$

Given,

                        $\Delta H = 77.2 KJ/mol;$

                        $\Delta S = 122 J{ K }^{ -1 }{ mol }^{ -1 };$

and Temperature = 400 K

                        $= 77200 - 400\left( 122 \right)$

                        $= 28400 J/mol$

                        ${ \Delta G }^{ \circ } = -2.303RT \log { K }$

Here, R = 8.314; T = 400 K; $\Delta H = 28400 J/mol$

                        $\log { K } = -28400 \left( 2.303 \times 8.314 \times 400 \right) = 4.292$

                        $K = 1.96 \times { 10 }^{ -4 }$

Answer 2

We know that,

Given,

\Delta H = 77.2 KJ/mol;ΔH=77.2KJ/mol;

\Delta S = 122 J{ K }^{ -1 }{ mol }^{ -1 };ΔS=122JK

−1

mol

−1

;

and Temperature = 400 K

= 77200 - 400\left( 122 \right)=77200−400(122)

= 28400 J/mol=28400J/mol

{ \Delta G }^{ \circ } = -2.303RT \log { K }ΔG

∘

=−2.303RTlogK

Here, R = 8.314; T = 400 K; \Delta H = 28400 J/molΔH=28400J/mol

\log { K } = -28400 \left( 2.303 \times 8.314 \times 400 \right) = 4.292logK=−28400(2.303×8.314×400)=4.292

K = 1.96 \times { 10 }^{ -4 }K=1.96×10

−4