Given : Kp = 9 x 10-10
Temperature = T = 273 + 25 = 298 K
ΔG0 = ?
ΔG0 = -2.303 RTlog10Kp
= -2.303 × 8.314 × 298 × log109 × 10-10
= -2.303 × 8.314 × 298 × [\(\bar{10}.9542\)]
= – 2.303 × 8.314 × 298 × [ – 9.0458]
= 51683 Jmol-1
= 51.683 kJmol-1
∴ ΔG0 = 51.683 kJmol-1