![\Delta G=-675.38 \frac{kJ}{mol}](/tpl/images/1200/7338/cab45.png)
Explanation:
Hello!
In this case, for this problem, it is possible to use the thermodynamic definition of the Gibbs free energy:
![\Delta G=\Delta H-T\Delta S](/tpl/images/1200/7338/f7eb0.png)
Whereas G, H and S can be assumed as constant over T; thus, we can calculate H at 135.4 °C:
![\Delta H=\Delta G+T\Delta S\\\\\Delta H=-775.41\frac{kJ}{mol}+(135.4+273.15)K*(0.81791\frac{kJ}{mol*K} )\\\\\Delta H=-441.58\frac{kJ}{mol}](/tpl/images/1200/7338/ca961.png)
Now, we can calculate the Gibbs free energy at 12.7 °C as shown below:
![\Delta G=-441.58\frac{kJ}{mol} -(12.7+273.15)K*0.81791\frac{kJ}{mol*K}\\\\\Delta G=-675.38 \frac{kJ}{mol}](/tpl/images/1200/7338/ebcb3.png)
Best regards!