Which of the following is increased by decreasing the volume of the reaction system in the following reaction: $\ce{2H_2S(g) +3O_2(g)<=> 2SO_2(g) +2H_2O(g)} + {\text{heat}}$

I. Rate of Reaction

II. Equilibrium concentration of reactants

III. Value of $\ce{K_{eq}}$

My attempt: By the gas laws, decreasing the volume of the container will increase the pressure, so equilibrium should shift to the right, as there are less gases on the right. Thus III is true. Additionally, heat is also generated by the shifting of the equilibrium, and the increased pressure on the system due to the decreasing of volume would also speed up the reaction, so I is also true. II is not true because the Equilibrium concentration of the products increases, not the reactants. So my final answer is I and III. Yet the answer key says I only. Why is that?


4 Answers 4


I think a good place to start is with the idea that that $K_{eq}$ will not change in value unless the temperature changes (see the second answer here for a good explanation). With this in mind we can say that statement III is false because the temperature is not being changed.

Statement II is shown to be false via Le Chatelier's principle, as the equilibrium concentration of reactants should decrease rather than increase due to the decrease in volume because the products have fewer gaseous moles and so would exert less pressure on the container.

Statement I can be shown to be true via Le Chatelier's principle by the same logic as Statement II: for the reaction to shift towards the product, the rate of formation of product must increase.

Therefore, only Statement I is true.

You might wonder how $K_{eq}$ can stay constant, but the rate of forward reaction can increase. Its important to remember that $K_{eq}$ is the ratio of the forward and reverse rate constants, not the forward and reverse rates. The rate constants have no dependence on volume (or pressure) so they won't change.

  • $\begingroup$ I am having some confusions. If the $K_{eq}$ remains constant, the rate of backward reaction should too increase in proportion with the forward. So why we observe more formation of the products? $\endgroup$ May 1, 2017 at 15:34

Few links that are relevant :

As given in the first link $K_{eq}$ will change only if you change temperature. What changes here is reaction quotient, $Q$.

$Q$ is calculated just like $K$ : $$\frac{(concentration\, of\, products)^a}{(concentration\, of\, reactants)^b}$$ where a and b are the respective stoichiometric coefficients.
It is clear that when you will decrease the volume of the container you will increase the concentration of both the reactants and products. But $Q$ will decrease as $\ce{O_2}$ in the reactants is cubed while both the products are squared.

If the system has to attain equilibrium $K$ has to be equal to $Q$ thus $Q$ has to increase which means that the concentration of the reactants has to decrease and that of the products has to increase.
Therefore the reaction shifts in the forward direction and the rate of the reaction increases.
Thus only I is right.



You should base it on the amount of moles on both sides

so if you wanted to figure it out u know that the side with the more moles is going to be affected i.e. if you decrease the pressure the side with the more moles of gas is going to decrease and shift the other way

to clarify:

1) You said that I is true but it isn't because we have to understand that only temperature changes the rate of the reaction so for rate of reaction we don't care about the decrease in volume.

2)For equilibrium concentration of reactants, we know that there is 5 moles of gases on the left side and there is four moles on the right side with heat, because you are not chaining the heat, when you decrease the volume that means that you are forcing the moles from the left to the moles on the right side.

3)For $K_{eq}$ we will go with your explanation of why it is true. But then we can rule out number 1. But it seems to me that your textbook either has a typo that tried to say "which of these are not true" or that you have not read the problem correctly.

  • 1
    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$
    – Mithoron
    Apr 11, 2017 at 17:29
  • $\begingroup$ @Mithoron. fixed $\endgroup$
    – John Rawls
    Apr 11, 2017 at 18:29
  • $\begingroup$ I'm gonna click 'Looks OK' in review, but it would be nice if you elaborate a bit. Answers here are expected to be as elaborate as possible. If you edit, notify me and I'll consider upvoting your answer. $\endgroup$
    – M.A.R.
    Apr 11, 2017 at 18:50
  • $\begingroup$ Chem SE as a site is trying to move away from giving hints as answers. If you want to just leave a hint, you can leave it as a comment. Answers need to be more fleshed out and give all the information to solve a problem (though that doesn't necessarily mean doing the whole problem step by step). $\endgroup$
    – Tyberius
    Apr 11, 2017 at 20:37
  • $\begingroup$ @M.A.R. It is fixed $\endgroup$
    – John Rawls
    Apr 11, 2017 at 21:45

Mind well, equilibrium constant $K_c$ as well as $K_p$ depends only on the temperature, they remain unaffected even though pressure is changed, or concentration of reactant or product is changed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.