Question

For a particular reaction at 234.1 °C, ΔG = -832.94 kJ/mol, and ΔS = 923.32 J/(mol·K). Calculate ΔG for this reaction at -105.7 °C.

Answer 1

We will employ the equation relating ΔG, ΔH and ΔS as

ΔG = ΔH – T.ΔS where T is the temperature of the system in Kelvin.

For our first part, we will use the supplied values of ΔG and ΔS to estimate ΔH.

The temperature of the system is 234.1 °C, or in the absolute scale,

T = (273 + 234.1) K = 507.1 K.

We have to convert the supplied value of ΔS in kJ. Since 1 J = 10^-3 kJ, hence, we have,

ΔS = 923.32 J/(mol.K) * (10^-3 kJ)/(1 J) = 0.92332 kJ/ (mol.K)

Substituting in the equation for ΔG, we obtain,

ΔG = ΔH – T.ΔS

or, (-832.94 kJ/mol) = ΔH – (507.1 K).(0.92332 kJ)/ (mol.K)

or, (-832.94 kj/mol) = ΔH – (468.22 kJ/mol)

or, ΔH = (-832.94 kJ/mol) + (468.22 kJ/mol)

or, ΔH = -364.72 kJ/mol

Now, we will use these values of ΔH and ΔS at temperature (-105.7°C) to estimate ΔG.

The experimental temperature in the absolute scale is

T = {273 + (-105.7)} K = 167.3 K

We simply substitute values (point to note is that we assume ΔG to be temperature dependent and ΔH and ΔS temperature independent).

ΔG = ΔH – T. ΔS

or, ΔG = (-364.72 kJ/mol) – (167.3 K).(0.92332 kJ)/(mol.K)

or, ΔG = (-364.72 kJ/mol) – (154.47 kJ/mol)

or, ΔG = -210.25 kJ/mol

The value of ΔG at -105.7°C is -210.25 kJ/mol. (ans)