Chemistry - How can the equilibrium shift, while Kc remains constant?

Solution 1:

$K_c$ 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.)

Solution 2:

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:

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