Question

For a particular reaction at 152.8 °C, ΔG = -966.13 kJ/mol, and ΔS = 459.28 J/(mol·K). Calculate ΔG for this reaction at -54.9 °C.

Answer 1

To solve this question, assume that ΔH and ΔS do not depend on temperature.   

Given, ΔG = -966.13 kJ/mol = -966130 J/mol

ΔS = 459.28 J/(mol·K)

T = 152.8 °C = (152.8+273) K = 425.8 K

Using the below equation, we get the ΔH for the reaction

ΔG = ΔH - TΔS ...........(1)

Putting the above values in equation (1), we get

=> -966130 J/mol = ΔH - (425.8 K)x459.28 J/(mol·K)

=> ΔH = -966130 + 195561.42 = -770568.58 J/mol = -770.569 kJ/mol

Now, take the value of ΔH and ΔS from above and we can calculate the ΔG at -54.9 °C

T = -54.9 °C = (-54.9 + 273) K = 218.1 K

ΔG = -770568.58 J/mol - (218.1 K)x459.28 J/(mol·K

= -770568.58 J/mol - 100168.97 J/mol

= -870737.548 J/mol

ΔG = -870.734 kJ/mol

Hence, ΔG for this reaction is -870.734 kJ/mol at -54.9 °C.