Consider the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Fe}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Fe}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) $$ Calculate \(\Delta G^{\circ}\) for the reaction. \(K_{\mathrm{sp}}\) for \(\mathrm{Fe}(\mathrm{OH})_{2}\) is \(1.6 \times 10^{-14}\).

Short Answer

Expert verified
After carrying out the required calculations, we find that the Standard Gibbs free energy, \(\Delta G^{\circ}\), for the reaction is approximately \(172 kJ/mol\).

Step by step solution

01

Convert the given temperature from \(\mathrm{C}\) to \(\mathrm{K}\)

The temperature is given in Celsius degrees. The conversion from Celsius to Kelvin is done by adding 273 to the Celsius temperature, hence \(T = 25^{\circ} C + 273 = 298 K\)
02

Insert the known values into the equation

The equation used to relate the standard Gibbs free energy, \(\Delta G^{\circ}\), and the equilibrium constant, \(K_{sp}\), is: \(\Delta G^{\circ} = -RT \ln K_{sp}\). We know that \(R = 8.314 J/(mol·K)\), \(T = 298 K\), and \(K_{sp} = 1.6 \times 10^{-14}\). Inserting the known values in the equation, we get \(\Delta G^{\circ} = -8.314 J/(mol·K) * 298 K * \ln(1.6 \times 10^{-14})\)
03

Calculate the value of \(\Delta G^{\circ}\)

Carrying out the computation in the equation from step 2, the value of \(\Delta G^{\circ}\) is obtained.

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