# Finding the equilibrium concentration of a reactant 'A'

We have the following reaction:- $$A+\ce{B<=>2C}+D$$ The initial concentration of A and B are 1M each.$K_c$ is $10^8$.Find the equilibrium concentration of A.

This is what I have tried:-

The initial concentration of A,B,C,D are 1,1,0,0 respectively The final concentrations of A,B,C,D have to be (1-$\alpha$),(1-$\alpha$),$2\alpha$,$\alpha$ respectively.

$$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$ $$10^8=\frac{(2\alpha)^2\alpha}{(1-\alpha)^2}$$ $$10^8(1-\alpha)^2=4\alpha^3$$ But even on solving this cubic equation on a calculator, the values are incorrect.

No need for the cubic equation. Just firstly assume that A and B reacted completely, then [C]=2M, [D]=1M, and from Kc you can calculate [A] which is the same as [B]. That's the approximation usually used with this kind of problems (the answer should be $2*10^{-4}M$).

Your approach seems correct. What is the "correct" numerical value of the answer according to your source?

I plugged in your equation to Wolfram Alpha, but using $x$ instead of an explicit $10^8$ as the factor on the left-hand side.

Since it is a cubic equation, there are (in general) three roots. In chemistry problems that involve cubic equations, usually two of the roots will violate physical principles of the problem. In this example, the physically meaningful answer must (i) be real, i.e. have zero imaginary part, and (ii) satisfy $0 \le y \le 1$.

Numerical noise can make exactly satisfying those constraints, especially point (ii), somewhat difficult. You will note that the approximate answers supplied by Wolfram Alpha are:

• $$0.999368 - 1.164153 \times 10^{-9} i$$
• $$1.00063 - 1.047738 \times 10^{-9} i$$
• $$2,499,998$$

The extent of reaction must be between 0 and 1, so obviously the answer of ~2.499 million is not physical. The second answer has (a real part) of 1.00063, which is also higher than one and thus not physical. Thus, only the first answer of 0.999368 is physical. You may be wondering about the $- 1.164153 \times 10^{-9} i$ term, i.e. the imaginary part of the answer. It arises from numerical noise in the root-finding algorithm. It can be safely ignored.

Thus, the numerical answer for $y$ is $\approx 0.999368$. That means that the equilibrium concentration of $A$ is $(1-y)^2$, or $3.99 \times 10^{-7}$ molar.

This makes chemical sense, as we expect that reactions which have very large equilibrium constants will go very nearly to completion.

• my answer section says that the answer has to be $2 \times 10^4$. I'm hoping that's a printing error. – Abhishek Mhatre Oct 13 '15 at 5:37
• If you start off with one molar each of A and B, then that is two molar total. The reaction forms three moles from two, so the maximum concentration of anything formed by the reaction is 3. So 20,000 is certainly wrong. – Curt F. Oct 13 '15 at 11:10