# How can the equilibrium shift, while Kc remains constant?

Consider the following reversible reaction.

$$\ce{Cr2O7^2-(aq) + H2O(l) <=> 2 CrO4^2-(aq) + 2 H+(aq)}$$

What will happen to the position of equilibrium and the value of $$K_c$$ when more $$\ce{H+}$$ ions are added at constant temperature?

$$\begin{array}{lcc} \hline & \text{Position of equilibrium} & \text{Value of}~K_c \\ \hline \text{A.} & \text{shifts to the left} & \text{decreases} \\ \text{B.} & \text{shifts to the right} & \text{increases} \\ \text{C.} & \text{shifts to the right} & \text{does not change} \\ \text{D.} & \text{shifts to the left} & \text{does not change} \\ \hline \end{array}$$

The answer given is D.

I understand that when more $$\ce{H+}$$ ions are added, the reaction shifts in reverse. So I've understood that this is known as having the equilibrium shift to the left, toward the reactants.

What confuses me is why $$K_c$$ doesn't change. If the equilibrium shifts to the left, I suppose that must mean more reactants are formed in relation to the products, so $$K_c$$ would decrease.

But the answer key says that $$K_c$$ remains unchanged. This is coherent with what I've been taught, but I don't understand how the equilibrium can change yet have $$K_c$$ remain constant.

I'm just afraid I've got some of the theory mixed up, so I wanted to get some clarity. Could someone explain this please?

$$K_c$$ is related to the ratio of reactants to products at equilibrium.

If the reaction is currently at equilibrium, and you add more products then the reaction is now out of equilibrium and the reverse reaction will happen until it is back in equilibrium.

I don't like the wording of "equilibrium shifts to the left" myself, I would say that reverse reactions occurs to restore equilibrium. But $$K_c$$ doesn't change based on reactants/products concentrations since its the ratio at equilibrium. $$K_c$$ will normally depend on temperature though.

Based on the options you have, $$K_c$$ not changing and the reverse reaction occurring, the closest answer would be D. (If you take equilibrium shifting to the left to mean the reaction goes in that direction.)

• Agreed. $K_c$ is the equilibrium constant. It does not do us much good if it changes when you add stuff! May 13, 2013 at 11:12
• Agreed. One should recognize that when more products are added, the new concentration of products then decreases by reaction, the concentration of the reactants increase, and the equilibrium ratio is unchanged. May 25, 2013 at 21:04

To add to Nick's great answer, I believe you are confusing $K_c$, the equilibrium constant, with $Q_c$, the reaction quotient.

The reaction quotient is defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, under any conditions or relative amounts of the various species.

$$Q_\mathrm c =\dfrac{[\ce{CrO4^{2-}}]^2[\ce{H+}]}{[\ce{Cr2O7^{2-}}][\ce{H2O}]}$$

The equilibrium constant describes which combinations of reactants and products constitute equilibrium, and there exists an infinite set of such combinations at a given temperature. However, they all must have $Q_\mathrm c = K_\mathrm c$. The difference in the formula for $K_\mathrm c$ below (from the formula for $Q$ above) is the use of equilibrium concentrations. $$K_\mathrm c= Q_\mathrm {c,eq} =\frac{[\ce{CrO4^{2-}}]^2_{\text{eq}}[\ce{H+}]_{\text{eq}}}{[\ce{Cr2O7^{2-}}]_{\text{eq}}[\ce{H2O}]_{\text{eq}}}$$

Let's take an example, where $K_\mathrm c=1$. There are infinitely many concentration combinations that we could have that would work. Three of them are shown below. $$1. \ \ \ [\ce{CrO4^{2-}}]=[\ce{H+}]=[\ce{Cr2O7^{2-}}]=[\ce{H2O}]=\pu{1 M}$$ $$2. \ \ \ [\ce{CrO4^{2-}}]=\pu{4 M};\ \ \ \ [\ce{Cr2O7^{2-}}]=[\ce{H2O}]=\pu{2 M}; \ \ \ \ [\ce{H+}]=\pu{1 M}$$ $$3. \ \ \ [\ce{CrO4^{2-}}]=[\ce{Cr2O7^{2-}}]=\pu{1 M};\ \ \ \ [\ce{H2O}]=[\ce{H+}]=\pu{0.987 M}$$

If we take the first case, where all concentrations are equal, and increase the concentration of $\ce{H+}$ by $\pu{0.5 M}$, now we have the following: $$[\ce{CrO4^{2-}}]=[\ce{Cr2O7^{2-}}]=[\ce{H2O}]=1\ M; \ \ \ \ [\ce{H+}]=\pu{1.5 M}$$

If we calculate $Q_c$, we find that $Q_c=1.5$. Thus, the system is no longer at equilibrium because $Q_c\neq K_c$.

$$Q_c=\frac{(\pu{1 M})^2(\pu{1.5 M})}{(\pu{1 M})(\pu{1 M})}=1.5$$

The system will shift towards the reactants (since $Q_\mathrm c>K_\mathrm c$) to reestablish equilibrium. A little algebra, and we can figure out just how much. The product concentrations will decrease and the reactant concentrations will increase, each by $nx$, where $x$ is the concentration change and $n$ is the stoichiometric coefficient.

$$[\ce{CrO4^{2-}}]_{\text{eq}}=1-2x\ M; \ \ \ \ [\ce{H+}]_{\text{eq}}=1.5-x\ M\ \ \ \ [\ce{Cr2O7^{2-}}]_{\text{eq}}=[\ce{H2O}]_{\text{eq}}=1+x\ M$$

$$K_\mathrm c=\frac{(1-2x\ M)^2(1.5-x\ M)}{(1+x\ M)(1+x\ M)}=1$$ $$K_\mathrm c=\frac{(1-4x+4x^2)(1.5-x)}{(1+2x+x^2)}=\frac{(1.5-7x+10x^2-16x^3)}{(1+2x+x^2)}=1$$ $$(1+2x+x^2)>0\ \ \text{for}\ \ x>0$$ $$1.5-7x+10x^2-16x^3=1+2x+x^2$$ $$0=16x^3-9x^2+9x-0.5$$

The above expression has three roots, two of which are imaginary. The real root is $x=0.0586353$. Thus, the new equilibrium concentrations are:

$$[\ce{CrO4^{2-}}]_{\text{eq}}=0.883\ M; \ \ \ \ [\ce{H+}]_{\text{eq}}=1.441\ M\ \ \ \ [\ce{Cr2O7^{2-}}]_{\text{eq}}=[\ce{H2O}]_{\text{eq}}=1.059\ M$$

• Let me see if I understand this correctly: so, if I perform a certain modification within the reaction mixture, my Q will automatically change, but then one of the equilibrium's reaction will become "more prominent" (faster in terms of rate) than the other so as to reverse the process and make Q = Kc again? Apr 3 at 17:00

See if we have $$K_c = \pu{100 mol dm-3}$$ and the concentration of product changes to $$\pu{50 mol}$$ so the net change of concentration is $$50$$ means half of the concentration decrease and if we look at initial concentration so it was for $$\pu{1 dm3}$$ and we always talk about $$\pu{1 dm3}$$ so to get the same value of $$K_c$$ you will have also to divide the volume by $$2$$ means now the value of $$K_c$$ is as under

$$K_c = \frac{50}{\frac{1}{2}} = \pu{100 mol dm-3}$$

Because volume also change.