Consider the following reaction at 298 K . 3 2 O 2 g ⇌ O 3 g . K p = 2 . 47 × 10 - 29 ∆ r G 0 for the reaction is _________ kJ . (Given R = 8 . 314 JK – 1 mol – 1 ) Round off your answer to the nearest integer.

Step-by-step solution

Given R = 8 . 314 JK – 1 mol – 1 3 2 O 2 g ⇌ O 3 g . K p = 2 . 47 × 10 - 29 Standard Gibbs free energy of the formation of a compound is basically the change of Gibbs free energy that is followed by the formation of one mole of that substance from its component element available at their standard states or the most stable form of the element which is at 25 ° C and 100 kPa. Its symbol is Δ f G ˚ . ∆ r G 0 = - RT ln K p = - 8 . 314 × 10 – 3 × 298 × ln ( 2 . 47 × 10 – 29 ) = – 8 . 314 × 10 – 3 × 298 × ( – 65 . 87 ) =