Question

(4) calculate $\triangle_{\text{r}}g^{\theta}_{\text{m}}$, judge the spontaneity
$\ce{2 co(g) + o_{2}(g) = 2co_{2}(g)}$
$\triangle_{\text{f}}g^{\theta}_{\text{m}}(\text{kj}\cdot\text{mol}^{-1})$ -137.168 0 -394.359

Explanation:

Step1: Recall the formula for reaction Gibbs free energy

The formula for the standard molar reaction Gibbs free energy \(\Delta_{r}G_{m}^{\ominus}\) is \(\Delta_{r}G_{m}^{\ominus}=\sum
u_{i}\Delta_{f}G_{m,i}^{\ominus}\), where \(
u_{i}\) is the stoichiometric coefficient of species \(i\) (positive for products, negative for reactants) and \(\Delta_{f}G_{m,i}^{\ominus}\) is the standard molar formation Gibbs free energy of species \(i\).

For the reaction \(2\mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) = 2\mathrm{CO}_{2}(\mathrm{g})\), the stoichiometric coefficients are: \(
u_{\mathrm{CO}}=- 2\), \(
u_{\mathrm{O}_{2}}=-1\), \(
u_{\mathrm{CO}_{2}} = 2\).

Step2: Substitute the values into the formula

We know \(\Delta_{f}G_{m}^{\ominus}(\mathrm{CO},\mathrm{g})=- 137.168\space\mathrm{kJ\cdot mol^{-1}}\), \(\Delta_{f}G_{m}^{\ominus}(\mathrm{O}_{2},\mathrm{g}) = 0\space\mathrm{kJ\cdot mol^{-1}}\), \(\Delta_{f}G_{m}^{\ominus}(\mathrm{CO}_{2},\mathrm{g})=-394.359\space\mathrm{kJ\cdot mol^{-1}}\).

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Step3: Judge the spontaneity

A reaction is spontaneous at standard conditions if \(\Delta_{r}G_{m}^{\ominus}<0\). Since \(\Delta_{r}G_{m}^{\ominus}=- 514.382\space\mathrm{kJ\cdot mol^{-1}}<0\), the reaction is spontaneous.

Answer:

\(\Delta_{r}G_{m}^{\ominus}=-514.382\space\mathrm{kJ\cdot mol^{-1}}\), and the reaction is spontaneous.