IMAT 2020 Chemistry Q47

Practice Question Solving
1

Why is answer C favoured in comparison to A even though they essential have the same outcome?

P, Q and R represent substances involved in a gaseous reaction.
The equilibrium constant (Kc) and the sign of the enthalpy change for the reaction are:

Kc= (P)^2/(R)(Q)^3

Assuming that all other conditions remain constant, a change in which factor results in an increase of the value of the equilibrium constant, K​c?

  1. A an increase in pressure
  2. B use of a suitable catalyst
  3. C a decrease in temperature
  4. D a decrease in pressure
  5. E an increase in temperature
2

hi! for this question what i thought about was the Pv=nRT formula. yes in normal cases if the pressure is increased while keeping other conditions normal, then the equilibrium will shift to the fewer gas moles side. but in this case, as we don’t keep the temperature the same, in the case of increased pressure, the temperature will also increase in the gases. this will cause the reaction to not be fully shifted as the reaction itself is exothermic. that’s why a decrease in the pressure will cause a much more big effect on the shift, increasing the Kc.

3

Right okay, thank you.

1 Like
4

wait isn’t the answer is C not A? It’s because Kc is not affected by pressure. Although pressure shifts the equilibrium it won’t change the equilibrium constant. Only temperature affects the equilibrium constant.

1 Like
5

Yes you are correct, I had forgotten that Kc is only affected by temperature. And the answer is indeed C

2 Likes
6

yes the answer is C indeed. i was just telling how i made sense out of it but i forgot that K can be only changed by temperature, you are right too

7

Ah okay I was confused by your explanation as you said that “that’s why a decrease in the pressure will cause a much more big effect on the shift, increasing the Kc.” when pressure would not increase the Kc :)) that’s why

8

ah sorry i think i was really tired and explained a bit flawed… thanks for telling mee