Consider the following reaction at 298 K. 2(g) 3(g) P 3 O O .K 2.47 10 . rG for the reaction is ____ — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

2(g) 3(g) P 3 O O .K 2.47 10 .    rG  = – RT ln KP = – 8.314 × 10–3 × 298 × ln (2.47 × 10–29) = –8.314 × 10–3 × 298 × (–65.87) = 163.19 kJ