Question

The following equilibrium is established at 273 K in a 5.0 L container

2NO_(g) + O_2(g) « 2NO_2(g)                   ΔH⁰ = - 113 kJ / mol

State and explain, using differential rates i.e. what happens to R_forwards and R_backwards, the effects the following have on the system at equilibrium – DO NOT use LCP as an explanation as LCP does not explain anything, it just helps you predict what will happen.

1. Selectively removing some of the NO_(g)
2. Increasing the temperature
3. Decreasing the volume of the container

Answer 1

When selectively NO is removed , reaction moves in direction to create more NO. Thus R forwards is higher than Rbackwards ( as forward reaction creates more NO)

dHo = -113 KJ/mol. Here heat is released in reaction. If we increase the temperature reaction will move in direction such that heat is absorbed and temperature is lowered. Heat absorbed for backwards reaction as forward reaction shows heat released. Thus increasing tempertaure makes Rbackwards higher than Rforward.

Decreasing volume is nothing but inrcreasing pressure. Now reaction will move in direction to reduce pressure. So reaction will proceed in direction where there are less gas molecules. Products side has less gas molecules ( 2 gas molecules compared to reactants where 3 gas molecules present). Hence reaction proceeds forward such that pressure is lowered or volume is increases. Thus Rforward is higher than R backwards