= Solution
Use the <pressure correction to a mineral reaction> for the solids, with $P_0=1\,\mathrm{bar}$. The <water> contribution must instead use its <chemical fugacity>. For pure <solid> phases and <water vapour> treated as an <ideal gas>, $f_{\mathrm{H_2O}}=P$, giving
$$
0=\Delta_rH^\circ-T\Delta_rS^\circ
+\Delta V_{\rm solids}(P-P_0)+RT\ln(P/P_0).
$$
The last term already accounts for the gas-pressure contribution; adding another gas-volume correction would double count it. The pressure ratio inside the logarithm is dimensionless.
With pressure expressed in bars, the solid correction is $-2.28(4500-1)=-10257.72\,\mathrm{J\,mol^{-1}}$. Thus
$$
\boxed{T=\frac{130862.28}{170.7-8.3145\ln4500}
=1298.755\,\mathrm K\simeq1025.605\,{}^\circ\mathrm C.}
$$
Using $4500$ rather than $4499$ for the solid pressure increment would silently shift the specified one-bar reference.
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