Calculate ΔrG∘ at 298 K for the following reactions. CO (g) + H2O (g) →H2 (g) + CO2 (g) 2-Predict the effect on ΔrG∘ of lowering the temperature for the reaction above.ΔrG∘ will decrease with decreasing temperature.ΔrG∘ will increase with decreasing temperature.ΔrG∘ will change slightly with decreasing temperature.

1) ΔG°r (298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

CO (g) + H2O (g) → H2 (g) + CO2 (g)

1) ΔG°r = ∑νiΔG°f, i

⇒ ΔG°r (298 K) = ΔG°CO2 (g) + ΔG°H2 (g) - ΔG°H2O (g) - ΔG°CO (g)

from literature, T = 298 K:

∴ ΔG°CO2 (g) = - 394.359 KJ/mol

∴ ΔG°CO (g) = - 137.152 KJ/mol

∴ ΔG°H2 (g) = 0 KJ/mol ... pure substance

∴ ΔG°H2O (g) = - 228.588 KJ/mol

⇒ ΔG°r (298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol) - ( - 137.152 KJ7mol)

⇒ ΔG°r (298 K) = - 28.619 KJ/mol

2) K = e∧ (-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K. mol

∴ T = 298 K

⇒ K = e∧ (-28.619 / (8.314 E-3) (298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r (200 K) = - (8.314 E-3) (200) Ln (9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r (298 K) = - 28.619 KJ/mol