# Question #cb815

May 22, 2015

Here's how you'd figure this out. First of all, your first equilibrium reaction is actually

$2 S {O}_{3 \left(g\right)} r i g h t \le f t h a r p \infty n s 2 S {O}_{2 \left(g\right)} + {O}_{2 \left(g\right)}$ $\textcolor{b l u e}{\left(1\right)}$

Now take a look at the second equilibrium reaction.

$S {O}_{2 \left(g\right)} + \frac{1}{2} {O}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s S {O}_{3 \left(g\right)}$ $\textcolor{b l u e}{\left(2\right)}$

Notice that this time, sulfur trioxide, $S {O}_{3}$ is on the reactants' side, as opposed to the first equilibrium reaction in which it was on the products' side.

Moreover, the stoichiometric coefficients are halved for all the species that take part in the reaction, i.e. you have 1 mole of $S {O}_{3}$ instead of 2 moles, 1 mole of $S {O}_{2}$ instead of 2 moles, and 1/2 moles of ${O}_{2}$ instead of 1 mole.

So, start by writing the equilibrium constant for the second equilibrium, ${K}_{c 2}$

${K}_{c 2} = \frac{\left[S {O}_{3}\right]}{\left[S {O}_{2}\right] \cdot {\left[{O}_{2}\right]}^{\text{1/2}}}$

The equilibrium constant for the first reaction, ${K}_{c 1}$, will be

${K}_{c 1} = \frac{{\left[S {O}_{2}\right]}^{2} \cdot \left[{O}_{2}\right]}{{\left[S {O}_{3}\right]}^{2}}$

You need to figure out how to write ${K}_{c 1}$ using ${K}_{c 2}$. The first thing you need to do is get sulfur trioxide in the denominator, since the first equilibrium features $S {O}_{3}$ as a reactant, not as a product. Flip ${K}_{c 2}$ to get

$\frac{1}{K} _ \left(c 2\right) = \frac{\left[S {O}_{2}\right] \cdot {\left[{O}_{2}\right]}^{\text{1/2}}}{\left[S {O}_{3}\right]}$

Now you need to get the exponents to match those of ${K}_{c 1}$. Notice that if you raise $\frac{1}{K} _ \left(c 2\right)$ to the power of 2, you get

$\frac{1}{{K}_{c 2}} ^ 2 = \frac{{\left[S {O}_{2}\right]}^{2} \cdot {\left[{O}_{2}\right]}^{\frac{1}{2} \cdot 2}}{{\left[S {O}_{3}\right]}^{2}} = \frac{\left[S {O}_{2}\right] \cdot \left[{O}_{2}\right]}{{\left[S {O}_{3}\right]}^{2}}$

But you know that

$\frac{{\left[S {O}_{2}\right]}^{2} \cdot \left[{O}_{2}\right]}{{\left[S {O}_{3}\right]}^{2}} = {K}_{c 1}$

Therefore,

$\frac{1}{K} _ {\left(c 2\right)}^{2} = {K}_{c 1} \implies {K}_{c 1} = \textcolor{g r e e n}{\frac{1}{49} ^ 2}$