Question

For a particular reaction at 135.4
°C, Δ=−775.41 kJ/mol
, and Δ=817.91 J/(mol⋅K)
.

Calculate ΔG for this reaction at 12.7
°C.

Answer 1

Answer:

$\Delta G=-675.38 \frac{kJ}{mol}$

Explanation:

Hello!

In this case, for this problem, it is possible to use the thermodynamic definition of the Gibbs free energy:

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

Whereas G, H and S can be assumed as constant over T; thus, we can calculate H at 135.4 °C:

$\Delta H=\Delta G+T\Delta S\\\\\Delta H=-775.41\frac{kJ}{mol}+(135.4+273.15)K*(0.81791\frac{kJ}{mol*K} )\\\\\Delta H=-441.58\frac{kJ}{mol}$

Now, we can calculate the Gibbs free energy at 12.7 °C as shown below:

$\Delta G=-441.58\frac{kJ}{mol} -(12.7+273.15)K*0.81791\frac{kJ}{mol*K}\\\\\Delta G=-675.38 \frac{kJ}{mol}$

Best regards!