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

`\Delta G=-675.38 (kJ)/(mol)`

Step-by-step 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(kJ)/(mol)+(135.4+273.15)K*(0.81791(kJ)/(mol*K) )\\\\\Delta H=-441.58(kJ)/(mol)`

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

`\Delta G=-441.58(kJ)/(mol) -(12.7+273.15)K*0.81791(kJ)/(mol*K)\\\\\Delta G=-675.38 (kJ)/(mol)`

Best regards!