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The equilibrium constant for the following reaction at $\pu{600^{\circ}C}$ is 4.0. Initially, two moles of $\ce{CO}$ and one mole of $\ce{H2O}$ were mixed in a $\pu{1.0 L}$ container. Determine the concentration of all species at equilibrium.

Attempt #1:

$$\ce{CO (g) + H2O (g) -> CO2 (g) + H2 (g)}$$

\begin{array}{|r|c|c|c|} \hline \text{Initial}~(M) & 2.0 & 1.0 & 0 & 0 \\ \hline \text{Change}~(M) & -x & -x & +x & +x \\ \hline \text{Equilibrium}~(M) & 2.0 – x & 1.0 - x \\ \hline \end{array}

Plug $x$ into the equilibrium expression and solve for $x$. $4.0 = 2[x]$, so $x = 0.85 [1.0 – x][2.0 – x]$.

Determine concentrations: the equilibrium values become 2.0 – 0.85, 1.0 - 0.85, 0.85, and 0.85, giving 1.2, 0.1, 0.85, and 0.85.

Where does the 0.85 come from? Could it be cross multiplied some way or is there another way this is done?

Furthermore, I want to make sure that I've done the 3.00 correctly for $\ce{Fe(SCN)2+}$.

Consider the reaction represented by the equation: $$\ce{Fe^3+ (aq) + SCN- (aq) -> Fe(SCN)2+ (aq)}$$

Attempt #2:

\begin{array}{ccc} \text{Initial} & \pu{6.00 M}\ \ce{Fe^{3+} (aq)} & \pu{6.00 M}\ \ce{SCN^{−} (aq)} \\ \text{Equilibrium} & \pu{? M}\ \ce{FeSCN2^+ (aq)}, \ K = 0.33 \end{array}

\begin{align} \frac{1}{3}x^2 - \frac{124}{25}x + \frac{297}{25} &= 0 & x_{(1;2)} &= \frac{\color{red}{\pmb{-}}(-\frac{124}{25})\pm\sqrt{\left(\frac{124}{25}\right)^2 - 4\cdot\frac13\frac{297}{25}}}{\frac{2}{3}} \\ && x_{(1;2)} &= \frac{124/25\pm\sqrt{\frac{15376}{(625)}-\frac{15601}{1000}}}{\frac{2}{3}} \\ && &= \left(124/25\pm{3}\right)\cdot\frac{3}{2} \\ && x_{(1;2)} &= 12;3 \end{align}

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The text

4.0 = [x]2 so x = 0.85 [1.0 – x][2.0 – x]

appears to be an error. The proper equation is:

$$4 = \frac{x^2}{(2-x)(1-x)}$$

The 0.85 is one of two roots of the quadratic equation you'll solve, using

$$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$

after you simplify the first expression, above.

As for the rest of your question, please expand: What does

the 3.00

mean?

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  • $\begingroup$ For that question, you've provided initial concentrations of the species on the left hand side. There is information that is missing in order to complete the problem. If you provide an equilibrium concentration for the right hand side, we can then compute the equilibrium constant (K). Alternatively, you can provide the equilibrium constant and we can then back-out what the concentrations of all the species are at equilibrium. Please provide either K or the concentration of the species on the right hand side at equilibrium. $\endgroup$ – Todd Minehardt Jun 22 '15 at 23:20

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