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At $25 ^\circ C$, $K=0.090$ for the reaction $$\ce{H2O (g) + Cl2O (g) -> 2HOCl (g)}$$ calculate the concentrations of all species at equilibrium for the case: $\pu{1.0 g} \ce{H2O}$ and $\pu{2.0 g} \ce{Cl2O}$ are mixed in a $\pu{1.0 L}$ flask.

I don't know where I go wrong in my calculations:

We have $\pu{0.0555 M}$ $\ce{H2O}$ and $\pu{0.023 M}$ $\ce{Cl2O}$. So at equilibrium, \begin{align} \ce{[H2O]} &= 0.0555 - x,\\ \ce{[Cl2O]} &= 0.023 - x\\ \ce{[HOCl]} &= 2x\\ \therefore K &= \frac{[\ce{HOCl}]^2}{[\ce{H2O}][\ce{Cl2O}]} \end{align}

After some moving about I get to $$-3.91x^2 - 7.1\times10^{-3}x + 1.15\times10^{-4} = 0$$ Where I end up getting $x = -8.74\times10^{-3}$ or $x=8.4\times10^{-3}$.

And this is where it all goes wrong, doing some quick calculations, I see that it does not match what the book says the answer is. So I'm doing something wrong, but I don't know what it is or where it's at.

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migrated from physics.stackexchange.com Nov 30 '13 at 8:40

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I suggest you to use this schematic: \begin{array}{l*{5}{c}}\hline \text{Equation} & \ce{H2O} & \ce{Cl2O} & \ce{<=> 2 HClO}\\[2ex]\hline \text{Initial concentration} /\pu{M} & \pu{0.055 mol//L} & \pu{0.023 mol//L} & 0 \\[2ex] \text{Equilibrium variation} /\pu{M} & - x & -x & +2x \\[2ex] \text{Equilibrium concentration} /\pu{M} & \pu{0.055 M} - x & \pu{0.023 M} - x & 2x \\[2ex] \hline \end{array}

At this point you have to substitute into the equilibrium equation you mentioned: \begin{align} 0.090 &= \frac{(2x)^{2}}{(0.055 - x)(0.023 - x)}\\ 0.090 &= \frac{4x^{2}}{0.001265 - 0.055x - 0.023x + x^2}\\ 4x^2 &= 0.090 (0.001265 - 0.055x - 0.023x + x^2)\\ 4x^2 &= 0.090 (0.001265 - 0.078x + x^2)\\ 4x^2 &= 0.000114 - 0.007x + 0.09x^2\\ 0 &=3.91x^2 + 0.007x - 0.000114\\ \end{align} So the positive result is: $$ x_2 = \frac{-0.007 + \sqrt{4.9\cdot10^{-5} + 0.00178}}{7.8}\approx 0.0046$$

An then equilibrium concentration are for $\ce{[H2O]} = 0.055 - 0.0046 = 0.050$, $\ce{[Cl2O]} = 0.018$ and $\ce{[HClO]} = 2x = 0.092$. This are my results.

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