'1 and 2 only' is the correct answer, since an equilibrium is defined as being closed (1) and having equal rate of reaction in both directions (2), but the concentrations of reactants and products do not need to be the same (3), and rarely are. The third statement might better read 'the concentrations of reactants and products remain constant', which they do, but not the same as each other.

The equilibrium constant is defined as being the product of the concentrations of the products (each to the power of the number of moles) over the product of the concentrations of the reactants (each to the power of the number of moles), e.g. for a reaction:

aA + bB ⇌ cC + dD

The value of K_c is 0.044. This number is less than 1, and indicates that the bottom of the fraction is larger than the top. This means that there are more reactants than products in the equilibrium mixture, and this is what 'lies to the left' means.

Thus the correct answer is 'The equilibrium lies to the left; there are more reactants than products in the equilibrium mixture'.

Remember that K_c is constant at constant temperature. So addition of a catalyst will not affect K_c.

Addition of a catalyst will not affect the position of an equilbrium either. It will decrease the time it takes for a reaction to reach equilibrium, because it will speed up the rate of the foward and backward reactions equally, but will not alter the position of the equilibrium when it gets there.

Increasing temperature will favour the endothermic direction, which is the reverse direction in this case (a negative ΔH indicates an exothermic forward reaction). The equilibrium will shift to the left to give a greater proportion of reactants. A greater proportion of reactants will make the bottom of the fraction greater and will therefore reduce the value of K_c.

The value of K_c will be < 0.044.

The correct answer is 'Q is the same as the K_c expression, but not necessarilly with equilibrium concentration values'.

Q is useful when a reaction is thrown out of equilibrium e.g.

A + B ⇌ C

\(K_c = {{[C]} \over [A][B]}\) at equilibrium, and

\(Q = {{[C]} \over [A][B]}\)

The reaction is at equilibrium (K_c = Q), and then more A is added: The reaction is suddenly thrown out of equilibrium and now Q will be smaller than K_c (since [A] has increased and it is on the bottom of the fraction). The equilibrium will now shift to the right, decreasing the amount of A (and B) and increasing the amount of C, until Q once more equals the value of K_c. Equilibrium is re-attained.

The correct answer is 'Q will initially decrease; the equilibrium will then shift causing Q to increase again until Q = K_c'.

\(K_c = {{[NO_2]^2} \over [N_2O_4]}\) ; this is also the expression for Q.

The reaction is at equilibrium (K_c = Q), and then more N₂O₄ is pumped in: The reaction is suddenly thrown out of equilibrium and now Q will be smaller than K_c (since [N₂O₄] has increased and it is on the bottom of the fraction). The equilibrium will now shift to the right, decreasing the amount of N₂O₄ and increasing the amount of NO₂, until Q once more equals the value of K_c. Equilibrium is re-attained.

Remember that K_c is constant at constant temperature.

The value of K_c will therefore remain as 0.044 when the volume decrease causes an increase in pressure, but the increase in pressure will favour the side of the equilibrium with fewest moles of gas (2 moles of the right) so the equilibrium will shift right.

Decreasing the volume (and thus increasing total pressure) causes the concentration values to each increase. There are more concentration terms on the bottom than the top of the fraction, so the bottom becomes bigger; decreasing the volume throws the reaction out of equilibrium and Q becomes smaller than K_c. Thus the equilibrium shifts to increase the top of the fraction (more ammonia, NH₃); that is a shift to the right, until Q once more equals K_c (0.044).

\(Q = {{[NH_3]^2} \over [N_2][H_2]^3}\)