Question

Which of the following reactions will have the largest value of K at 298 K? A) CaCO3(s) → CaO(s) + CO2(g) ΔG° =+131.1 kJ B) 2 Hg(g) + O2(g) → 2 HgO(s) ΔG° = ‒180.8 kJ C) 3 O2(g) → 2 O3(g) ΔG° = +326 kJ D) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔG° = ‒28.0 kJ

Answer 1

Answer :

(a) The value of 'K' for the reaction is $1.05\times 10^{-23}$  

(b) The value of 'K' for the reaction is $4.93\times 10^{31}$

(c) The value of 'K' for the reaction is $7.17\times 10^{-58}$

(d) The value of 'K' for the reaction is $8.09\times 10^{4}$

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs free energy

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K = equilibrium constant

Now we have to calculate the value of 'K' for the following reactions.

(a) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$    $\Delta G^o=+131.1kJ=+131100J$

$\Delta G^o=-RT\times \ln K$

$+131100J=-(8.314J/K.mol)\times (298K)\times \ln K$

$K=1.05\times 10^{-23}$

Thus, the value of 'K' for the reaction is $1.05\times 10^{-23}$

(b) $2Hg(g)+O_2(g)\rightarrow 2HgO(s)$    $\Delta G^o=-180.8kJ=-180800J$

$\Delta G^o=-RT\times \ln K$

$-180800J=-(8.314J/K.mol)\times (298K)\times \ln K$

$K=4.93\times 10^{31}$

Thus, the value of 'K' for the reaction is $4.93\times 10^{31}$

(c) $3O_2(g)\rightarrow 2O_3(g)$    $\Delta G^o=+326kJ=+326000J$

$\Delta G^o=-RT\times \ln K$

$+326000J=-(8.314J/K.mol)\times (298K)\times \ln K$

$K=7.17\times 10^{-58}$

Thus, the value of 'K' for the reaction is $7.17\times 10^{-58}$

(d) $Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)(s)$    $\Delta G^o=-28.0kJ=-28000J$

$\Delta G^o=-RT\times \ln K$

$-28000J=-(8.314J/K.mol)\times (298K)\times \ln K$

$K=8.09\times 10^{4}$

Thus, the value of 'K' for the reaction is $8.09\times 10^{4}$