= Solution
At the one-bar <standard state>, pure <mineral endmembers> and <water vapour> treated as an <ideal gas> all have unit <thermodynamic activity>. The <Gibbs energy of reaction> is $\Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ$. Setting it to zero gives
$$
\boxed{T=\frac{141120}{170.7}=826.714\,\mathrm K\simeq553.564\,{}^\circ\mathrm C.}
$$
The quantities are taken per mole of the reaction as written. The simplified result uses the stipulated temperature-independent <enthalpy> and <entropy> differences.
Back to article page
