Question

The heat of formation of MgO at 300 K and 1 bar pressure is -600.60 kJ mol$^{-1}$. The free energy (in kJ mol$^{-1}$) of formation of MgO at 280 K is ......... (Round off to nearest integer)

Hint:

When calculating free energy changes, remember to account for both the enthalpy and entropy changes over the temperature range.

Answer 1

Correct Answer: -570 - -574

To find the Gibbs free energy of formation of MgO at 280 K, we use the relationship:

ΔG = ΔH - TΔS

1. Calculate ΔH at 280 K:
The heat capacity data allows us to adjust the ΔH value from 300 K to 280 K using:

ΔH(T₂) = ΔH(T₁) + ∫(T₁ to T₂) ΔC_P dT

Given constant ΔC_P from 280 to 300 K, the ΔH change from 300 K to 280 K is:

ΔH(280) = ΔH(300) - ΔC_PΔT

where ΔC_P = C_P(MgO) - [C_P(Mg) + 0.5C_P(O₂)]

ΔC_P = 27.0 - [24.9 + 0.5(29.4)] = -12.6 J mol^-1 K^-1

ΔH(280) = -600.60 - (-12.6)(20/1000)

ΔH(280) = -600.60 + 0.252 = -600.348 kJ mol^-1

2. Calculate ΔS_reaction:
The entropy change for the reaction:

ΔS = S_m(MgO) - [S_m(Mg) + 0.5S_m(O₂)]

ΔS = 0 - [0 + 0.5(205.2)] = -102.6 J mol^-1 K^-1 = -0.1026 kJ mol^-1 K^-1

3. Calculate ΔG at 280 K:

ΔG(280) = ΔH(280) - TΔS

ΔG(280) = -600.348 - 280(-0.1026)

ΔG(280) = -600.348 + 28.728 = -571.62 kJ mol^-1

Final result: ΔG = -572 kJ mol^-1 (rounded to nearest integer)