Answer:
(a) See below
(b) (i) No effect; (ii) no effect
Step-by-step explanation:
(a) At equilibrium
XY ⇌ X + Y
I: 10 0 0
C: -x +x +x
E: 10-x x x
Kc = {[X][Y]}/[XY] = 0.11
(x × x}/(10 - x) = 0.11
x²/(10 - x)= 0.11
Check for negligibility:
10/0.11 = 91 < 400. x is not negligible.
We must solve a quadratic.
x²/(10 - x) = 0.11
x² = 0.11(10 - x) Distribute 0.11
x² = 1.1 - 0.11 x Collect all terms on left-hand side
x² + 0.11 x -1.1 = 0 Solve the quadratic
x = 0.995 ≈ 1
At equilibrium
[XY] = 10 - x = 10 - 1 = 9
[X] = x = 1
[Y] = x = 1
We have nine molecules of XY and one each of X and Y. The diagram will look like Figure 1 below.
(b) Effect of a catalyst
(i) On position of equilibrium
Figure 2 shows the effect of a catalyst.
A catalyst has no effect on the position of equilibrium.
The relative concentration of reactants and products will not change when a catalyst is added.
The catalyst provides an alternate reaction pathway with a lower Eₐ.
The molecules can get across the barrier faster, but the position of equilibrium will not change.
(ii) On ΔH
The addition of a catalyst does not affect the value of ΔH.
The molecules of reactants and products are the same, so ΔH does not change.
