# How do I calculate the equilibrium constant of a reaction from two related reactions' constants?

At a certain temperature the following reactions have the constants shown:

\begin{align}\ce{S(s) + O2(g) &<=> SO2(g)} && K'_\mathrm{c} = 4.2\times 10^{52}\\ \ce{2S(s) + 3 O2(g) &<=> 2 SO3(g)} && K''_\mathrm{c} = 9.8\times 10^{128}\end{align}

Calculate the equilibrium constant $K_\mathrm c$ for the following reaction at that temperature:

$$\ce{2SO2(g) + O2(g) <=> 2 SO3(g)}$$

I already have the answer (which is given in the back of the textbook): $$K_\mathrm{c}=5.6\times10^{23}$$

But whenever I complete the problem (I've done it thrice) I get $2.3\times10^{76}$.

My process for solving this is as follows. I reverse the first equation and double the coefficients. Since I've done this, I use the inverse of $K'_\mathrm{c}$, $(K_\mathrm{c}')^{-1}$. I then multiply both $K$ values and the result is $2.3\times10^{76}$. Where is my mistake?

• when you double the coefficients you don't get double the $K_{eq}$ but the square of $K_{eq}$. – MaxW Mar 3 '17 at 2:09

Equilibrium constants depend upon the choice of reaction equations. I have noted the extended way to calculate $K''_\mathrm{c}$ below:

\begin{align}\ce{2S + 3 O2 &<=> 2 SO3}\tag{2}\\[0.4em] K''_\mathrm{c} &= \frac{[\ce{SO3}]^2}{[\ce{O2}]^3}\end{align}

(Sulfur does not take part in the equilibrium constant because its activity, as a solid, is $a(\ce{S}) = 1$.)

Having this, we can approach the question. First, we need to add particles in an equivalent way (i.e. the same to reactant and product side) to get from equation $(3)$ to something that can be expressed as a linear combination of $(1)$ and $(2)$ ($a(1) + b(2)$):

\begin{align}\ce{2SO2 + O2 &<=> 2SO3}\tag{3}\\[0.4em] \ce{2SO2 + 3 O2 &<=> 2 SO3 + 2 O2}\tag{$3'$}\\[0.4em] \ce{2 SO2 + 3 O2 + 2 S &<=> 2 SO3 + 2 O2 + 2 S}\tag{$3''$}\end{align}

$\displaystyle (3'') = -2 \times (1) + 1 \times (2)$

With that in hand, we can determine what the equilibrium constant of $(3'')$ would look like and how that can be traced back to the other equilibrium constants:

$$\displaystyle K_\mathrm{c} = \frac{[\ce{SO3}]^2[\ce{O2}]^2}{[\ce{SO2}]^2[\ce{O2}]^3}\tag{4}$$
$$\displaystyle K_\mathrm{c} = \frac{[\ce{O2}]^2}{[\ce{SO2}]^2} \times \frac{[\ce{SO3}]^2}{[\ce{O2}]^3} = (K'_\mathrm{c})^{-2} \times K''_\mathrm{c} \tag{5}$$

And thus, your final solution is:

$$(4.2\times10^{52})^{-2} \times (9.8\times10^{128}) = 5.6\times10^{23}\tag{6}$$