Molarity Numerical on Water Gas Shift Reaction


> 
> For the water gas shift reaction below, $K\_c = 3.491$ at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if $\pu{0.3815 mol}$ of $\ce{CO}$ and $\ce{H2O}$ are initially mixed in a $\pu{250 mL}$ flask?
> 
> 
> $$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$$
> 
> 
> 


I was able to calculate the molarity of $\ce{CO}$ and $\ce{H2O}$ and set up an ICE table, but my answer seemed inaccurate. My work so far:


$$3.491 = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]}$$


$$3.491 = \frac{x^2}{(1.526-x)^2}$$


Because $x$ should be much smaller than $1.5$, it can be ignored when squaring $1.526$. Therefore, $3.491 = x^2/2.32$


$$3.491\cdot 2.32 = x^2$$


The square root of that value is $2.85$, which is impossible because there weren't $\pu{2.85 mol}$ of $\ce{CO}$ or $\ce{H2O}$ to begin with.




Taking the square root of both sides gives


$$\frac{x}{1.526-x}=1.868$$


So


$$x=1.526\frac{1.868}{2.868}=0.994$$


 1

Your reasoning looks correct up to point you assumed $x\ll 1.5$.
There is no evidence to support this claim and eventually you just have to solve the equation:


$$K\_c = \frac{x^2}{(c\_0 - x)^2}\label{eqn:1}\tag{1}$$


where $c\_0$ is the initial concentration of the reactants:


$$c\_0 = \frac{n\_0}{V} = \frac{\pu{0.3815 mol}}{\pu{0.250 L}} = \pu{1.526 M} \tag{2}$$


Equation \eqref{eqn:1} can be rewritten as a typical quadratic equation of $ax^2 + bx + c = 0$ form:


$$(K\_c - 1)x^2 - 2K\_cc\_0x + K\_cc\_0^2 = 0 \tag{3}$$


with the following set of roots:


$$
\begin{align}
x\_{1,2} &= \frac{2K\_cc\_0 ± \sqrt{4K\_c^2c\_0^2 - 4K\_cc\_0(K\_c - 1)}}{2(K\_c - 1)} \\
 &= \frac{K\_cc\_0 ± \sqrt{K\_cc\_0(K\_cc\_0 - c\_0(K\_c - 1))}}{K\_c - 1} \\
 &= \frac{5.327 ± 2.851}{2.491}\tag{4}
\end{align}
$$


$$x\_1 = 3.283;\quad x\_2 = 0.994$$


Since $x < c\_0$, only $x\_2 = 0.994$ is physically meaningful, resulting in the final answer:


$$[\ce{CO2}] = [\ce{H2}] = x = \pu{0.994 M}$$


$$[\ce{CO}] = [\ce{H2O}] = c\_0 - x = \pu{1.526 M} - \pu{0.994 M} = \pu{0.532 M}$$


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