Question

Can someone please help?

Answer 1

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.