Equilibrium constant Kc =
(1) 2 NO(g) + O2(g) 2NO2(g) ∆H = -65KJ/mol-1
The forward reaction is an exothermic reaction meaning it’s giving of heat. The reverse reaction is endothermic if the temperature is increased.
(2) N2(g) + O2(g) 2NO(g)
In this reversible reaction, I think that if the concentration is lowered, then the position of the equilibrium would be shifted to the left. If the concentration was to increase, then the position would shift to the right.
(3) S(g) + O2(g) SO2 ∆H = +265KJ/mol-1
The position of the equilibrium when it is heated moves to the left because there is more heat produced