Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g) + Br2 (g) A) adding NOBr. B) removing NO C) doubling the concentration of NO at the same time as the concentration of Br2 is cut in half. D) cutting the concentrations of both NOBr and NO in half

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr (g) ↔ 2NO (g) + Br₂ (g)

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question, the changes that will not affect the equilibrium will be, reducing the concentration of both NOBr and NO to half,

Hence,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence,

K' = K