Ijust need on c-g. the whole question is there in case anyone needs it.

(c) 0.11; (d) -24.5 kJ·mol⁻¹; (e) See below; (f) Kc increases;

(g) No effect on ΔH

Step-by-step explanation:

(c) Kc at 125 °C  

In Part (a) one molecule of XY dissociated into one X and one Y.

       XY ⇌ X + Y

I:      10       0    0

C:     -1       +1   +1

E:      9        1     1

Kc = {[X][Y]}/[XY] = (1 × 1)/9 = ⅑ = 0.11

(d) ΔH

ΔH is a constant that is characteristic of the reaction.

ΔH = -24.5 kJ·mol⁻¹

(e) Effect of temperature on concentrations

If ΔH is negative, the reaction is exothermic.

Heat is a product of the reaction, so we can write the equation as

XY ⇌ X + Y + heat

If we lower the temperature, we are removing heat from the system.

Le Châtelier's Principle states that if you apply a stress to a system at equilibrium, it will respond by trying to relieve the stress.

We applied a stress by removing heat, so the system responds by producing more heat. The position of equilibrium moves to the right, and more products will form.

The diagram might look like the one below.

(f) Effect of temperature on Kc

Kc = [Products]/[Reactants]

We are increasing [Products] and decreasing [Reactants].

If you increase the numerator and decrease the denominator, you increase the value of the quotient.

The value of the equilibrium constant increases when the temperature decreases.

(g) Effect of temperature on ΔH

Decreasing the temperature has no effect on ΔH, because the enthalpies of the reactants and products are properties of the substances themselves. They do not depend on the temperature.

the answer is the