ΔG° for the formation of HI(g) from its gaseous elements is -2.303 kcal/mol at 500 K. When the partial pressure of HI is 10 atm and of I₂(g) is 0.001 atm, what must be the partial pressure of hydrogen

Question

ΔG° for the formation of HI(g) from its gaseous elements is -2.303 kcal/mol at 500 K. When the partial pressure of HI is 10 atm and of I₂(g) is 0.001 atm, what must be the partial pressure of hydrogen be at this temperature to reduce the magnitude of ΔG for the reaction to zero?

Prashant Kotian · Accepted Answer

Correct answer: A. For 1/2 H₂ + 1/2 I₂ ⇌ HI, ΔG° = -2303 cal/mol. ΔG = ΔG° + 2.303 RT log Q = 0. -2303 + 2.303(2)(500) log Q = 0 => -2303 + 2303 log Q = 0 => log Q = 1 => Q = 10. Q = P_HI / (P_H₂^0.5 × P_I₂^0.5) = 10 / (P_H₂^0.5 × 0.001^0.5) = 10. Thus, (P_H₂ × 0.001)^0.5 = 1 => P_H₂ × 0.001 = 1 => P_H₂ = 1000 atm. Therefore, correct answer is A.

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