Question

Kp for the following reaction is 0.16 at 25 degree C. 2 NOBr(g) 2 NO(g) Br_2(g) The enthalpy change for the reaction at standard conditions is

Answer 1

Step-by-step explanation:

Given that:

`2 NOBr_((g)) \iff 2 NO_((g)) + Br_(2(g))`

From above:

`K_p = 0.16 = ((P_(NO))^2 (P_(Br)))/((P_(NOBr))^2)`

To predict the effect of the addition of Br₂(g);

The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr

The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.