Calculate Kp for the following reaction at 500.0°C

Question

3NO(g) <--> N₂O(g) + NO₂(g)

using the free-energy data for the following reaction at the same temperature.

6NO(g) <--> 2N₂O(g) + 2NO₂(g)  ΔG° = −209.1 kJ

 

Answer 1

3NO(g) <--> N₂O(g) + NO₂(g)

Given that ... 6NO(g) <--> 2N₂O(g) + 2NO₂(g)  ΔG° = −209.1 kJ

Note that 3NO(g) <--> N₂O(g) + NO₂(g) is ONE HALF of 6NO(g) <--> 2N₂O(g) + 2NO₂(g)

When you divide an equation by a factor, you must also divide the ∆H by that same factor.

Thus...3NO(g) <--> N₂O(g) + NO₂(g) ∆H = 1/2 x -209.1 kJ

3NO(g) <--> N₂O(g) + NO₂(g) ∆H = 104.6 kJ