Molarity Numerical on Water Gas Shift Reaction

Question

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.

Answer 1

Taking the square root of both sides gives

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

So

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

Answer 2

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:

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

$$[\ce{CO}] = [\ce{H2O}] = c_0 - x = \pu{1.526 M} - \pu{0.994 M} = \pu{0.532 M}$$