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

(a) The value of 'K' for the reaction is
`1.05* 10^(-23)`

(b) The value of 'K' for the reaction is
`4.93* 10^(31)`

(c) The value of 'K' for the reaction is
`7.17* 10^(-58)`

(d) The value of 'K' for the reaction is
`8.09* 10^(4)`

Explanation :

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

`\Delta G^o=-RT* \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* \ln K`

`+131100J=-(8.314J/K.mol)* (298K)* \ln K`

`K=1.05* 10^(-23)`

Thus, the value of 'K' for the reaction is
`1.05* 10^(-23)`

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

`\Delta G^o=-RT* \ln K`

`-180800J=-(8.314J/K.mol)* (298K)* \ln K`

`K=4.93* 10^(31)`

Thus, the value of 'K' for the reaction is
`4.93* 10^(31)`

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

`\Delta G^o=-RT* \ln K`

`+326000J=-(8.314J/K.mol)* (298K)* \ln K`

`K=7.17* 10^(-58)`

Thus, the value of 'K' for the reaction is
`7.17* 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* \ln K`

`-28000J=-(8.314J/K.mol)* (298K)* \ln K`

`K=8.09* 10^(4)`

Thus, the value of 'K' for the reaction is
`8.09* 10^(4)`