Question

The value of ΔG° at 221.0°C for the formation of phosphorous trichloride from its constituent elements, P2(g) + 3Cl2(g) → 2PCl3(g) is ________ kJ/mol. At 25.0°C for this reaction, ΔH° is -720.5 kJ/mol, ΔG° is -642.9 kJ/mol, and ΔS° is -263.7 J/K.

Answer 1

ΔG = - 590.20 kJ/mol

Step-by-step explanation:

The formula for calculating Gibb's Free Energy can be written as:

ΔG = ΔH - TΔS

Given That:

ΔH = -720.5 kJ/mol

T = 221.0°C = (221.0 + 273.15) = 494.15 K

ΔS° = -263.7 J/K

So; ΔS° = -0.2637 kJ/K if being converted from joule to Kilo-joule

Since we are all set, let replace our given data in the above equation:

ΔG = (-720.5 kJ/mol) - (494.15 K) ( - 0.2637 kJ/K)

ΔG = (-720.5 kJ/mol) - (- 130.30755)

ΔG = -720.5 kJ/mol + 130.30755

ΔG = -590.192645 kJ/mol

ΔG = - 590.20 kJ/mol

Thus, The value of ΔG° at 221.0°C for the formation of phosphorous trichloride from its constituent elements, P2(g) + 3Cl2(g) → 2PCl3(g) is -590.20 kJ/mol.