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7th edition 5J.5 part b

Posted: Tue Jan 22, 2019 9:01 pm
by Ryan Danis 1J
This question asks whether reactants or products will be favored by an increase in pressure (resulting from compression) on the equilibria.

b. H2O(g)+C(s) <-> H2(g)+CO(g)

I understood parts a,c,d,e because when pressure increases, the side with the fewer number of gas molecules is favored because it minimizes the increase in pressure. However, there is an equal number of molecules on both the reactant and product side of part b. Why is the equilibria is effected (favors reactants)?

Re: 7th edition 5J.5 part b

Posted: Tue Jan 22, 2019 9:05 pm
by Samantha Ito 2E
You have to compare the number of moles of gas (don't include the solid).

Re: 7th edition 5J.5 part b

Posted: Wed Jan 23, 2019 2:10 pm
by Sophia Ding 1B
Thinking similarly to Kc or Kp, we don't include solids or liquids into them. So when determining how pressure will affect equilibrium, just like we would for Kp, we would not include the solid into consideration of equilibrium shift. Thus if pressure increases, we can use the shorthand rule of favoring the side with less molecules, which is the reactants.

Re: 7th edition 5J.5 part b

Posted: Wed Jan 23, 2019 4:11 pm
by Anna O 2C
The reactants are favored because in a closed vessel with solids and gases, the increase in pressure isn't going to affect the solid, it will only affect the gases. Because of this, increased pressure will cause the reaction to move towards the side with less moles of gas and thus less partial pressure. The reaction will move to the reactants so that the one mole of gas isn't affected as strongly as 2 moles of gas would be on the product side.