Remember that K_c is constant at constant temperature, but temperature is changing here so the value of K_c will change.

\(K_c = {[CO_2][H_2]\over{[CO][H_2O]}}\)

Increasing temperature will shift the equilibrium in the endothermic direction (to oppose the change in temperature in accordance with Le Chatelier's principle). The forward reaction is exothermic (negative ΔH) so the equilibrium will therefore shift left.

In the K_c expression, this will increase the quantity of reactants, increasing the denominator and therefore reducing the value of Kc.

The position of equilibrium will shift left and the value of K_c will decrease is therefore the correct answer.

Remember that K_c is constant at constant temperature, so the change in pressure will not affect K_c (although it will shift the position of equilibrium), however the change in temperature will affect K_c.

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

Decreasing temperature will favour the exothermic direction, which is the forward direction in this case (a negative ΔH indicates an exothermic forward reaction). The equilibrium will shift to the right to give a greater proportion of products (NH₃). A greater proportion of products will make the top of the fraction greater and will therefore increase the value of K_c.

Increases is therefore the correct answer.

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

Therefore the correct answer is:

\(K_c = {{[NOCl]^2} \over [NO]^2[Cl_2]}\)

As temperature is increasing here so the value of K_c decreases.

\(K_c = {[N_2O_4]\over{[NO_2]^2}}\)

That means that the bottom (denominator) of the fraction must be increasing, so the reaction must be shifting to the left; the reverse reaction is favoured as temperature increases.

The reverse reaction is favoured is therefore the correct answer.

Note that increasing the temperature will increase the rate of reaction in both directions, but not equally. As the reverse reation is favoured here, an increase in temperature will increase the rate of the reverse reaction to a greater extent than the rate of the forward reaction (that is how the equilibrium shifts).

K_c is for 2SO_2(g) + O_2(g) ⇌ 2SO_3(g) is:

\(K_c = {{[SO_3]^2} \over [SO_2]^2[O_2]}\)

K_c is for 4SO_2(g) + 2O_2(g) ⇌ 4SO_3(g) is:

\(K_c = {{[SO_3]^4} \over [SO_2]^4[O_2]^2}\)

It is important to know (law of indices) that when we multiply values with indices (powers) then we add the indices. Looking at the two K_c expressions we can see that in the second expression, all of the terms are a square of the terms in the first expression e.g. [SO₃]² squared is [SO₃]² × [SO₃]² = [SO₃]⁴.

As all the terms are squared, the equilibrium constant for the second equation must be the square of the equilibrium constant for the first equation.

K² is therefore the correct answer.

Addition of a catalyst will not affect the position of an equilbrium. A catalyst 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.

All the other factors will affect the position of equilibrium in this case; temperature always changes the position of equilibrium; change in pressure and change in volume will both affect the position of equilibrium as there are different numbers of moles of gas on each side of the equation.

Addition of a catalyst is therefore the correct answer.

If the pressure is decreased the equation will favour the side with greatest moles of gas (according to Le Chatelier's principle), which means the equilibrium will shift to the left (one mole of gas on the left and none on the right).

Shifting the equilibrium to the left will reduce the quantity of H⁺ ions in the aqueous solution, which will increase the pH (make it less acidic) as pH=−log₁₀[H⁺].

The reaction will shift to the left and the pH will increase is therefore the correct answer.

Adding a catalyst will never change the position of equilibrium (only increase the rate of attainment of the equilibrium position).

Increasing the temperature will favour the endothermic direction (according to Le Chatelier's principle) which is the left hand side; the forward reaction is exothermic as indicated by the negative ΔH.

Increasing the pressure will favour the side with fewest moles of gas (according to Le Chatelier's principle) which is the right hand side, with 2 moles compared to 3 moles on the left.

2 only is therefore the correct answer.

Addition of a catalyst will not affect the position of an equilbrium. A catalyst 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.

Addition of H⁺ (hydrogen) ions or SO₄²⁻ (sulfate) ions will add ions to the left hand side and according to Le Chatelier's principle will therefore shift the equilibrium to the right; favouring the forward reaction.

Addition of OH⁻ ions will cause the H⁺ ions to react: H⁺ + OH⁻ → H₂O, so will remove H⁺ ions from the left hand side and will therefore shift the equilibrium to the left; favouring the reverse reaction.

Addition of OH⁻_(aq) ions is therefore the correct answer.

The highest yield will be obtained by shifting the equilibrium to the right hand side (products).

The reaction is exothermic (negative ΔH), so a low temperature will favour the exothermic/forward direction and increase the yield of product.

There are fewer moles of gas on the right hand side of the equation, so a high pressure will favour the forward direction and increase the yield of product.

Low temperature and high pressure is therefore the correct answer.

(Note that these conditions will generate the highest yield, but may not be used in industry as a low temperature is likely to result in a slow rate of reaction.)

Remember that K_c is constant at constant temperature.

The value of K_c will therefore remain the same.

The decrease in volume 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.

Therefore The position of equilibrium will shift right, and the value of K_c will remain the same is the correct answer.

In explanation: 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 relatively bigger; thus 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 (K_c remains the same).

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