Find the stoichiometric coefficient of a reactant given Kp, Kc, and the stoichiometric coefficient of the product


> 
> Have an hypothetical equilibrium reaction:
> 
> 
> $$\ce{2 A (s) + x B (g) <=> 3 C (g)}$$
> 
> 
> With the constants
> 
> 
> \begin{align}
> K\_\mathrm{p} &= 0.0105 \\
> K\_\mathrm{c} &= 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?


$$\ce{x B(g) <=> 3 C (g)}$$


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.


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