Answer:
a) K ≅ 2.0 × 10⁹⁰
b) K = 2.0 × 10⁻²⁶
Explanation:
For the first reaction.
2 CO(g) + O₂(g) ⇄ 2CO₂(g)
Equilibrium constant equation be represented as:
[tex]\delta G^0[/tex] = -RT㏑K
㏑K = [tex]\frac{\delta G^0}{-RT}[/tex] ----------------- equation (1)
Where;
Gas Constant (R) = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹
Temperature (T) = 25°C = (25 + 273 K) = 298 K
Standard Gibbs free energy [tex]\delta G^0_{rxn}[/tex] = ???
To find the [tex]\delta G^0_{rxn}[/tex] for the above equation; we have:
[tex]\delta G^0_{rxn} =[2*\delta G^0_{f(co_2)}-(2*\deltaG^0_{f(co)}+1*\delta G^0_{f(o_2)})]kJ/mol[/tex]
[tex]\delta G^0_{rxn} =[2*394.4-(2*137.2+1*0)]kJ/mol[/tex]
[tex]\delta G^0_{rxn} = - 514.4 kJ/mol[/tex]
If we slot in our values back to equation (1); we have:
㏑K = [tex]-\frac{-514.4kJ/mol}{(8.314*10^-3kJmol^{-1}*(298K)}[/tex]
㏑K = 207.6
K = [tex]e^{207.6[/tex]
K = 1.5 × 10⁹⁰
K ≅ 2.0 × 10⁹⁰
Therefore, the equilibrium constants at 25 degrees Celsius for this reaction = 2.0 × 10⁹⁰
b)
For the second reaction.
2H₂S(g) ⇄ 2H₂(g) + S₂(g)
Equilibrium constant equation be represented as:
[tex]\delta G^0[/tex] = -RT㏑K
㏑K = [tex]\frac{\delta G^0}{-RT}[/tex] ----------------- equation (2)
Where;
Gas Constant (R) = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹
Temperature (T) = 25°C = (25 + 273 K) = 298 K
Standard Gibbs free energy [tex]\delta G^0_{rxn}[/tex] = ???
To find the [tex]\delta G^0_{rxn}[/tex] for the above equation; we have:
[tex]\delta G^0_{rxn} =[(2*\delta G^0_{f(H_2)}+1*\delta G^0_{f(S_2)})-2*\delta G^0_{f(H_2S)})]kJ/mol[/tex]
[tex]\delta G^0_{rxn} =[(2*0)+(1*79.7)-(2*-33.4})]kJ/mol[/tex]
[tex]\delta G^0_{rxn} =146.5kJ/mol[/tex]
Substituting our values, we have :
㏑K = [tex]-\frac{146.4kJ/mol}{(8.314*10^-3kJmol^{-1}*(298K)}[/tex]
㏑K = -59.13
K = [tex]e^{-59.13}[/tex]
K = 2.0 × 10⁻²⁶
Therefore, the equilibrium constants at 25 degrees Celsius for this reaction = 2.0 × 10⁻²⁶
