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[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

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...

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system
Under standard conditions and at $\mathrm{T}=298 \mathrm{~K}$, $\Delta G^{o}{ }_{298}$ can be calculated as the difference between the total Gibbs energy of formation of the reaction products and the total Gibbs energy of formation of the reactants. Necessary reference data: 
$$\Delta G_{f}^{0}(\mathrm{NaOH}, \tau)= -381.1 \, \text{kJ/mol}$$ 
$$\Delta G_{f}^{\circ}(\mathrm{Na}_{2} \mathrm{O}) = -378 \, \text{kJ/mol}$$ 
$$\Delta G_{f}^{0}\left(\mathrm{H}_{2} \mathrm{O}, \text{liquid}\right) = -237 \, \text{kJ/mol}.$$

Calculate $\Delta G^{\mathrm{o}}{ }_{298}$ using these values:
$$\Delta G^{\mathrm{o}}{ }_{298} = 2 \cdot \Delta G_{f}^{0}(\mathrm{NaOH}) - \left[ \Delta G_{f}^{0}\left(\mathrm{Na}_{2} \mathrm{O}\right) + \Delta G_{f}^{0}\left(\mathrm{H}_{2} \mathrm{O}\right) \right]$$
$$ = 2 \cdot (-381.1) - \left[ -378 + (-237) \right] = -147.2 \, \text{kJ}.$$

Given that the value of $\Delta G^{\circ}{ }_{298}$ is negative, the spontaneity of the reaction is possible.

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