Respuesta :
Explanation:
The reaction is as follows.
[tex]CuS \rightleftharpoons Cu(s) + \frac{1}{2}O_{2}(g)[/tex]
As, value of [tex]\Delta G[/tex] is positive. Therefore, reaction is non-spontaneous.
[tex]CuO(s) + C(graphite) \rightleftharpoons Cu(s) + CO(g)[/tex]
[tex]\Delta G = \sum \Delta G_{f}_{product} - \sum \Delta G_{f}_{reactant}[/tex]
= [-137.2 - (127.2 kJ/mol)]
= -10 kJ/mol
Since, value of [tex]\Delta G[/tex] is negative here so, reaction is spontaneous.
Also, [tex]\Delta G[/tex] = -RT ln
where, R = 8.314 J/mol K
T = [tex]400^{o}C[/tex] = (400 + 273) K = 673 K
K = equilibrium constant
[tex]-10 \times 10^{3} J/mol = -8.314 J/mol K \times 673 ln K[/tex]
100 = [tex]2.303 \times 8.314 J/mol K \times 673 log K[/tex]
log K = 0.00776
K = 1.018
Therefore, we can conclude that equilibrium constant for the coupled reaction is 1.018.
Answer:
[tex]CuO(s)<-->Cu(s)+\frac{1}{2}O_2(g);\Delta G^o_1=127.2kJ/mol \\2C(s)+O_2(g)<-->2CO(s);\Delta G^o_2=-137.16kJ/mol[/tex]
[tex]K=5.93[/tex]
Explanation:
Hello,
In this case, the coupled process contains the following two chemical reactions:
[tex]CuO(s)<-->Cu(s)+\frac{1}{2}O_2(g);\Delta G^o_1=127.2kJ/mol \\2C(s)+O_2(g)<-->2CO(s);\Delta G^o_2=-137.16kJ/mol[/tex]
By taking the total Gibbs free energy for this coupled reactions:
[tex]\Delta G^o_{T}=127.2kJ/mol-137.16kJ/mol=-9.96kJ/mol[/tex]
In such a way, we compute the equilibrium constant as follows:
[tex]K=exp(-\frac{\Delta G^o_{T}}{RT} )=exp(-\frac{(-9960J/mol)}{8.314J/molK*673.15} )\\K=5.93[/tex]
Best regards.