(Edexcel) I don't understand this question relating to Kp and reaction yields?

This is from Edexcel Unit 4, Jan 2010.

Q: Hydrogen is used in very large quantities as a fuel, as a reducing agent, and in the production of ammonia. Hydrogen is manufactured by steam reforming of methane from natural gas. Two reactions are involved, both being in equilibrium in closed
systems.
Reaction I CH4(g) + H2O(g) <---> CO(g) + 3H2(g) ∆H = + 210 kJ mol−1
Reaction II CO(g) + H2O(g) <---> CO2(g) + H2(g) ∆H = −42 kJ mol−1

(a) Write the expression for the equilibrium constant, Kp, for reaction I.

-- Kp =
p(H2)^3. p(CO)
p(CH4)p(H2O)

(b) Reaction I occurs at a temperature of 1000 K and a pressure of 30 atm over a nickel catalyst.

(i) State and explain the effect, if any, on the value of Kp of increasing the pressure
on the reaction.

-- No effect (as Kp dependent only on temperature)

Now, this part is what's confusing me:

(ii) Explain, in terms of your answers to (a) and (b)(i), why an increase in the pressure leads to a decrease in yield in reaction I.

Ans: First mark:

EITHER
mole fractions/partial pressures of numerator
decrease
OR
mole fractions/partial pressures of denominator
increase (1)
Second mark:

any mention of × PT^2
OR
xPT^4
PT^2 (1)

:confused:

So here PT is referring to the partial pressures (I guess). But I still don't get it, what does the "PT^2" have to do with an increase in the mole fraction?

Reply 1

EierVonSatan

I'm assuming that PT stands for total pressure...I will just use P for ease.

You've correctly stated the expression for K_p and if you break that down into mole fractions you get:

K_p = \frac{(X_{H2})^3 \times (X_{CO})}{(X_{CH4}) \times (X_{H2O})} \times \frac{P^4}{P^2}

The cube term on the top makes the difference. This part make sense?

So we can see that the numerator has become bigger than the denominator on the right hand side. Since K_p must remain constant, the left fraction must do the opposite to compensate. So the mole fractions on the top get smaller -> less reaction yield.