Find the stoichiometric coefficient of a reactant given Kp, Kc, and the stoichiometric coefficient of the product Have an hypothetical equilibrium reaction:
  
  $$\ce{2 A (s) + x B (g) &lt;=&gt; 3 C (g)}$$
  
  With the constants
  
  \begin{align}
K_\mathrm{p} &amp;= 0.0105 \\
K_\mathrm{c} &amp;= 0.45
\end{align}
  
  both at $\pu{250 ^\circ C}$. What is the value of $x$?


Since $\ce{A}$ is solid, I guess we can ignore it, right?

$$\ce{x B(g) &lt;=&gt; 3 C (g)}$$

Using $K_\mathrm{c}$:

$$0.0105 = \frac{\left[\ce{C}\right]^3}{\left[\ce{B}\right]^x}$$

Then

$$\left[\ce{B}\right]^x = \frac{\left[\ce{C}\right]^3}{0.0105}$$

But to solve this, I need the concentration of $\ce{C}$. How can I find it out?
Using the relation of $K_\mathrm{p}$ to $K_\mathrm{c}$:

$$\frac{K_\mathrm{p}}{K_\mathrm{c}} = (RT)^{\Delta n}$$

where $\Delta n$ is the change in the number of moles of gas in the reaction (here, $\Delta n = 3 - x$). Plugging in (using $R = \pu{0.0821 L atm mol-1 K-1}$ and $T = \pu{523.15 K}$) we get:

$$\frac{0.0105}{0.45} = 0.0233 = 42.950615^{(3-x)}$$

To solve we take the natural logarithm of both sides:

$$\ln{(0.0233)} = (3-x) \ln{(42.950615)}$$
$$\frac{\ln{(0.0233)}}{\ln{(42.950615)}} = -1 = (3 - x) \implies x = 4$$

And so $x = 4$. We can check to see if our answer is correct by plugging the calculated value for $x$ into the second expression from the top of this post:

$$0.0233 \buildrel{?}\over{=} 42.950615^{-1}$$

which does turn out to be correct.