Surely they’re not at equilibrium any more? I understand that a favourable condition can make a reaction in one direction go faster, but then surely either the forward or backward reaction is faster and the rates are no longer the same so there is no longer an equilibrium? I do gcse chemistry, I’m a bit confused :/

  • $\begingroup$ Reaction rate depend upon concentration. $\endgroup$
    – knoftrix
    Commented May 29, 2019 at 18:31
  • 2
    $\begingroup$ This is more like a figure of speech. This is still an equilibrium like the one you can observe for an asymmetrically balanced seesaw $\endgroup$
    – andselisk
    Commented May 29, 2019 at 18:37

3 Answers 3


What is equal at equilibrium?

The forward and reverse rates are equal at equilibrium.

What is constant at equilibrium?

The concentrations are constant, and the rates are constant.

What is typically not equal at equilibrium?

The concentrations of reactants and products are typically not equal at equilibrium. For a reaction with multiple species in solution or in the gas phase, there are multiple sets of conditions that are at equilibrium. With three or more species that can vary in concentration (i.e. not pure solid or pure liquid), concentrations don't have to be equal even if the equilibrium constant happens to equal one. So, equilibrium does not mean concentrations are equal.

[..] surely either the forward or backward reaction is faster and the rates are no longer the same so there is no longer an equilibrium?

A reaction can be at equilibrium when concentrations of reactants are different from concentrations of products because rate constants can be different too. For a simple (single-step) reaction $\ce{A(aq)<=>B(aq)}$, if the rate constant of the forward reaction is larger than that of the reverse reaction, then the product concentration has to be higher than the reactant concentration at equilibrium to make up for that.


Reaction rate depends on concentration of reactants.

Consider a simple reaction

$$\ce{A -> B} $$

In this reaction, rate in forward direction is given by rate law

$r_1 =k_1 \ce{[A] }$

And in backward direction

$r_2 =k_2\ce{[B]} $

Then eventually a time will come when $r_1=r_2$, since concentration of both A and B are either decreasing or increasing, if concentration of A increases then that of B will decrease or vice versa, (it is NOT necessary that $\ce{ [A] =[B]}$ at equibrium) at this time (when forward and backward rates are equal) the reaction is in equilibrium and no further products are formed, since both forward and backward rates are equal.

If you add catalyst to decrease activation energy of, say, forward reaction then forward rate $r_1$ will increase but concentration of $\ce{A}$ i.e., $\ce{[A]}$ is also decreasing and concentration of B is increasing so again equilibrium will be established when $r_1=r_2$ and we say that


I think your confusion is coming from the idea that the rate of the forward direction isn't permanently increased for example. As an example consider $$\ce{N2 + 3H2 <=>2NH3}$$ as this is a reaction you may be familiar with. If I had the reaction in a sealed container meaning an equilibrium will be established then the moles and thus concentration of each component will remain constant. If I then change a physical condition, say increase the pressure, the equilibrium position in the example will move towards the right - the rate of the forward reaction will increase. However just because the rate has increased the reaction will still reach a new equilibrium where the concentration of the species will remain constant. The equilibrium position will remain constant.

If you are interested maybe go ad read about equilibrium constants. for example the formation of ammonia will have an equilibrium constant $K_c$ which will remain constant regardless of any physical change made to the system, expect temperature. $$\ce{K_c}=\frac{[\ce{NH3}]^2}{\ce{[N2][H2]^3}}$$

  • $\begingroup$ So when people say the equilibrium has shifted to the right, does that just mean the concentration of the products has increased and not the rate? Thank you for your help by the way!! :) $\endgroup$ Commented May 29, 2019 at 19:28
  • $\begingroup$ Yeah exactly, the equilibrium position is the ratio of products to reactants so if the equilibrium position shifts right then the ratio of products to reactants increases. (we give the ratio in terms of concentrations). If this answered your question be sure to accept the answer too. $\endgroup$
    – H.Linkhorn
    Commented May 29, 2019 at 19:32
  • $\begingroup$ Ohhhhh! I get it now, thanks! And I just accepted it sorry, haven’t been on here in a while. $\endgroup$ Commented May 29, 2019 at 19:34
  • $\begingroup$ Thats great, glad you get it now $\endgroup$
    – H.Linkhorn
    Commented May 29, 2019 at 19:35
  • $\begingroup$ @H.Linkhorn In your answer you say that the equilibrium constant remains constant, but in your comment you say that the ratio of products to reactants increases. That is a contradiction. It is correct to say that when you change concentrations, there will be a net shift of the reaction to attain equilibrium again (with the same equilibrium constant, but a different set of equilibrium concentrations). $\endgroup$
    – Karsten
    Commented May 29, 2019 at 20:43

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