Finding equilibrium concentrations in solution from initial reagent concentrations and equilibrium constant

Question

We have the following reaction: $$\ce{A + B<=>2C +D}$$ The initial concentration of $\ce{A}$ and $\ce{B}$ are $\pu{1M}$ each. $K_c$ is $\pu{1.0E8}$. Find the equilibrium concentration of $\ce{A}$.

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. Therefore

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

Answer 1

No need for the cubic equation. Just firstly assume that $\ce{A}$ and $\ce{B}$ reacted completely, then $\ce{[C]}=\pu{2M}$, $\ce{[D]}=\pu{1M}$, and from $K_c$ you can calculate $\ce{A}$ which is the same as $\ce{B}$.

That's the approximation usually used with this kind of problem.

The answer should be $\pu{2E-4M}$.

Answer 2

This problem can be solved iteratively, if you are looking to increase precision. A first iteration ("zeroth order solution") was presented in another answer, where it was first assumed that approximately all reagent is converted to product before using the result to back-compute the residual (very small) concentration of reagent at equilibrium.

In fact you can rewrite the cubic equation into the following form:

$$\alpha = 1-\sqrt{\frac{4\alpha^3} {K_c}}$$

where $\alpha$ is the final concentration of $[\ce{D}]$ or the concentration of $[\ce{A}]$ consumed at equilibrium (as defined in the OP). From this definition we can write that the remaining concentration $\beta=[\ce{A}]_\textrm{fin}$ is

$$\beta = 1-\alpha$$

or

$$\beta_1= \sqrt{\frac{4(1-\beta_{0})^3} {K_c}}$$

This expression was used in another solution by using the approximation $\beta_0 \approx 0$. It can more generally be used to solve for $\beta$ iteratively starting from an initial guess (not necessarily $\alpha_0 =1$). For large $K$ convergence is pretty quick. In this case one iteration is enough ie if we set $\beta_0=0$ then

$$\beta_1=[\ce{A}]_\textrm{fin} = \pu{2.0E-4}$$

but even if we make a very bad initial estimate convergence is still pretty fast:

$$ \begin{array} {c|c} i & \beta_i\\ \hline 0 & 0.5 \\1 & \pu{7.0711e-5} \\2 & \pu{1.9998e-4} \\3 & \pu{1.9994e-4} \\4 & \pu{1.9994e-4} \end{array}$$

Answer 3

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 $-\pu{ 1.164153E-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 $\pu{3.99E-7}$ molar.

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