# How do you calculate the change in Gibbs free energy (∆G) for the following reactions at 25 °C?

## Ca(s) + 2H_2O(l) → Ca(OH)_2(s) + H_2(g) ∆H = -411.6 kJ/mol ∆S = 31.8 J/mol∙K?

Apr 25, 2016

You can do it like this:

#### Explanation:

Use this relationship which arises from the 2nd law of thermodynamics:

$\Delta G = \Delta H - T \Delta S$

$\Delta H$ is the enthalpy change

$T$ is the absolute temperature

$\Delta S$ is the entropy change

Putting in the values:

$\Delta G = - 411 \times {10}^{3} - \left(298 \times 31.8\right)$

$\Delta G = - 411 \times {10}^{3} - \left(9.476 \times {10}^{3}\right)$

$\Delta G = - 421.07 \text{ ""kJ/mol}$