So in class, I've learned about Equilibrium. So it is basically when the rate of the reverse and the forward reaction are the same. So I was like that sounds pretty easy, I can understand this. And then suddenly there was a twist (there always is in class). The teacher was like: so if you increase or add a stressor, the equilibrium will shift to stop it. How can equilibrium shift, I thought it was a concept not a "thing" to move up or down. Is it because it shifts to maintain the Equilibrium Constant?

I do not understand this. If this question has already been asked then please point me to that. I have even searched online and could not find an explanation in layman's terms.


1 Answer 1


Let's do a simple generic example. We have a reaction:

$$\ce{2A <=>[k_1][k_{-1}] B}$$

Experimentally we determine that the rates of the forward and reverse reactions are:

$$\text{Rate}_1=k_1[\ce{A}]^2$$ $$\text{Rate}_{-1}=k_{-1}[\ce{B}]$$

At equilibrium:

$$\text{Rate}_1=\text{Rate}_{-1}$$ $$k_1[\ce{A}]^2=k_{-1}[\ce{B}]$$

Since the rate constants are constants we can combine them into the equilibrium constant to generate the law of mass action:


Now, if $k_1=2k_{-1}$, we have a value of the equilibrium constant, which gives us the relationship between $[\ce{A}]_{eq}$ and $[\ce{B}]_{eq}$ that defines equilibrium:

$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{2k_{-1}}{k_{-1}}=2$$ $$K_{eq}=\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=2$$ $$2{[\ce{A}]_{eq}^2}=[\ce{B}]$$

In this case, only concentrations of $\ce{A}$ and $\ce{B}$ that satisfy $2{[\ce{A}]_{eq}^2}=[\ce{B}]$ constitute equilibrium. Other combinations of concentrations are not at equilibrium and therefore the forward and reverse rates are not equal.

Let's see how this works. An equilibrium situation is $[\ce{A}]=1\text{ M}$ and $[\ce{B}]=2\text{ M}$ because:


The rates are:

$$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(1^2)=2k_{-1}$$ $$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$

However, what if $[\ce{A}]=[\ce{B}]=2\text{ M}$?

$$\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=\frac{2}{2^2}=\frac{1}{2}\ne K_{eq}$$ $$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(2^2)=8k_{-1}$$ $$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$

The concentrations, when plugged into the mass action expression, do not match the relationship defined by the equilibrium constant. More importantly, the rates are no longer equal. In this example, the rate of the forward reaction is now 4 times the rate of the reverse reaction. This is how equilibrium is reestablished. The forward reaction is going faster, but as it does so, $\ce{A}$ is consumed (slowing down the forward reaction) and $\ce{B}$ is produced (speeding up the reverse reaction) until the rates are equal again.

We can even do a bit of algebra to figure out what the concentrations of $\ce{A}$ and $\ce{B}$ will be once equilibrium is reestablished. Let's have $x$ represent the change in concentration of $\ce{B}$, so the change in the concentration of $\ce{A}$ is $-2x$.

$$K_{eq}=2=\frac{[\ce{B}]}{[\ce{A}]^2}=\frac{2+x}{(2-2x)^2}=\frac{2+x}{4-8x+4x^2}$$ $$2(4-8x+4x^2)=2+x$$ $$8-16x+8x^2=2+x$$ $$6-17x+8x^2=0$$ $$x=0.44695..., 1.6781...$$

In our case, we must use $x=0.44695$ since the other value would lead to a negative concentration for $\ce{A}$. Thus:

$$[\ce{A}]_{eq}=2.00-2(0.44695)=1.11$$ $$[\ce{B}]_{eq}=2.00+0.44695=2.45$$

  • 1
    $\begingroup$ so what you are saying is that no matter what the stressor is, Equilibrium will return to maintain the equilibrium constant, either by alleviating either side of the equation. So just to make it clear, the equilibrium constant is essentially just a rate of the products over the reactants? Am I right. Its not about the concentrations but the rates. So what happens when equilibrium shifts, does the rate change? So what happens is like the equilibrium constant changes and then returns back to it. $\endgroup$
    – Asker123
    Commented Jan 6, 2015 at 16:42
  • 1
    $\begingroup$ Very close. Perturbing an equilibrium situation does disrupt the rate balance. The equilibrium constant is the ratio of the rate constants. The equilibrium constant does not change. What does change is the reaction quotient. For more information about the difference between equilibrium constant and reaction quotient, see chemistry.stackexchange.com/questions/4977/… $\endgroup$
    – Ben Norris
    Commented Jan 6, 2015 at 23:47

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