Finding equilibrium concentrations, trouble simplifying equation

Question

For $\ce{CO (g) + H2 (g) <=> CH2O (g)}$

with K_c = 0.068 @ 273K

initial concentrations being $\ce{[CO]}$ = 1.25M, $\ce{[H2]}$ = 2.00M, $\ce{[CH2O]}$ = 1.00M

I did $\frac{1.00}{((1.25)(2.00)} = 0.4$, which is more then K_c, so it's leaning to the left, will produce more reactants, and products will lower. So the I did:

$\frac{1.00 - x}{(1.25 + x)(2.00 + x)} = 0.068$

Multiplied the bottom and got $X^2 + 3.25x + 2.5$

so

$\frac{1.00 - x}{X^2 + 3.25x + 2.5} = 0.068$

I then tried multiplying the denominator by 0.068 and rearranging to solve for $x$ from there, but that's where I got stuck. $X$ is supposed to be 0.65.

Answer 1

Everything you do is correct and it is only the final algebraic transformation missing: \begin{align} 0.068 &= \frac{1-x}{(1.25+x)(2.00+x)}\\ 0.068 &= \frac{1-x}{x^2 + 3.25x + 2.5} &&|\cdot (x^2 + 3.25x + 2.5)\\ 1-x &= 0.068(x^2 + 3.25x + 2.5) &&|-1 +x\\ 0.0 &= 0.068x^2 + 1.221x -0.83\\ \end{align}

Now you can solve for $x$ according to \begin{align} 0&=ax^2+bx+c & x_{1,2}&=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\ \end{align}

And that gives me the values $x_1\approx0.656$ and $x_2\approx-18.6$. The latter one of course makes no sense.