Respuesta :
Answer :
(a) The value of 'K' for the reaction is [tex]1.05\times 10^{-23}[/tex]
(b) The value of 'K' for the reaction is [tex]4.93\times 10^{31}[/tex]
(c) The value of 'K' for the reaction is [tex]7.17\times 10^{-58}[/tex]
(d) The value of 'K' for the reaction is [tex]8.09\times 10^{4}[/tex]
Explanation :
The relation between the equilibrium constant and standard Gibbs free energy is:
[tex]\Delta G^o=-RT\times \ln K[/tex]
where,
[tex]\Delta G^o[/tex] = 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) [tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex] [tex]\Delta G^o=+131.1kJ=+131100J[/tex]
[tex]\Delta G^o=-RT\times \ln K[/tex]
[tex]+131100J=-(8.314J/K.mol)\times (298K)\times \ln K[/tex]
[tex]K=1.05\times 10^{-23}[/tex]
Thus, the value of 'K' for the reaction is [tex]1.05\times 10^{-23}[/tex]
(b) [tex]2Hg(g)+O_2(g)\rightarrow 2HgO(s)[/tex] [tex]\Delta G^o=-180.8kJ=-180800J[/tex]
[tex]\Delta G^o=-RT\times \ln K[/tex]
[tex]-180800J=-(8.314J/K.mol)\times (298K)\times \ln K[/tex]
[tex]K=4.93\times 10^{31}[/tex]
Thus, the value of 'K' for the reaction is [tex]4.93\times 10^{31}[/tex]
(c) [tex]3O_2(g)\rightarrow 2O_3(g)[/tex] [tex]\Delta G^o=+326kJ=+326000J[/tex]
[tex]\Delta G^o=-RT\times \ln K[/tex]
[tex]+326000J=-(8.314J/K.mol)\times (298K)\times \ln K[/tex]
[tex]K=7.17\times 10^{-58}[/tex]
Thus, the value of 'K' for the reaction is [tex]7.17\times 10^{-58}[/tex]
(d) [tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)(s)[/tex] [tex]\Delta G^o=-28.0kJ=-28000J[/tex]
[tex]\Delta G^o=-RT\times \ln K[/tex]
[tex]-28000J=-(8.314J/K.mol)\times (298K)\times \ln K[/tex]
[tex]K=8.09\times 10^{4}[/tex]
Thus, the value of 'K' for the reaction is [tex]8.09\times 10^{4}[/tex]
