Question

Heating copper(II) oxide at 400°C does not produce any appreciable amount of Cu: CuO(s) ⇆ Cu(s) + 1 2 O2(g) ΔG o = 127.2 kJ/mol However, if this reaction is coupled to the conversion of graphite to carbon monoxide, it becomes spontaneous. Write an equation for the coupled process and calculate the equilibrium constant for the coupled reaction. Be sure to include the states of the chemical species.

Answer 1

`CuO(s)<-->Cu(s)+(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`

`K=5.93`

Step-by-step explanation:

Hello,

In this case, the coupled process contains the following two chemical reactions:

`CuO(s)<-->Cu(s)+(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`

By taking the total Gibbs free energy for this coupled reactions:

`\Delta G^o_(T)=127.2kJ/mol-137.16kJ/mol=-9.96kJ/mol`

In such a way, we compute the equilibrium constant as follows:

`K=exp(-(\Delta G^o_(T))/(RT) )=exp(-((-9960J/mol))/(8.314J/molK*673.15) )\\K=5.93`

Best regards.

Answer 2

Step-by-step explanation:

The reaction is as follows.

`CuS \rightleftharpoons Cu(s) + (1)/(2)O_(2)(g)`

As, value of
`\Delta G` is positive. Therefore, reaction is non-spontaneous.

`CuO(s) + C(graphite) \rightleftharpoons Cu(s) + CO(g)`

`\Delta G = \sum \Delta G_(f)_(product) - \sum \Delta G_(f)_(reactant)`

= [-137.2 - (127.2 kJ/mol)]

= -10 kJ/mol

Since, value of
`\Delta G` is negative here so, reaction is spontaneous.

Also,
`\Delta G` = -RT ln

where, R = 8.314 J/mol K

T =
`400^(o)C` = (400 + 273) K = 673 K

K = equilibrium constant

`-10 * 10^(3) J/mol = -8.314 J/mol K * 673 ln K`

100 =
`2.303 * 8.314 J/mol K * 673 log K`

log K = 0.00776

K = 1.018

Therefore, we can conclude that equilibrium constant for the coupled reaction is 1.018.