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Question 17.p.6: Set up a reaction table and use the variables to find equili......

Set up a reaction table and use the variables to find equilibrium concentrations in the equilibrium expression.

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\text{The initial [HI]}=\frac{2.50\text{ mol}}{10.32\text{ L}} =0.242248 \ M  \\ \begin{matrix}  \text{Concentration }(M) & \text{2 HI(g)} & \leftrightarrows & \text{H}_2\text{(g)} & + & \text{I}_2\text{(g)} \\ \text{Initial} & 0.242248 && 0 & & 0 \\ \text{Change} & -2\text{x} && +\text{x} & & +\text{x} & \text{(2:1:1 mole ratio)} \end{matrix} \\ \begin{matrix} && \overline{\begin{matrix} \text{Equilibrium} & 0.242248-2\text{ x} &&& \text{x} & && &\text{x}& \end{matrix} }& \text{(unrounded values)} \end{matrix}

Set up equilibrium expression:

K_\text{c}=1.26\times 10^{-3}=\frac{\text{[H}_2]\text{[I}_2]}{\text{[HI}]^2} =\frac{\text{[x][x]}}{[0.242248-2\text{x}]^2}\quad\quad \text{Take the square root of each side} \\ 0.035496=\frac{\text{[x]}}{[0.242248-2\text{x}]} \\ 0.0085988-0.070992\text{x = x} \\ \text{x}=8.0288\times 10^{-3}=8.03\times 10^{-3} \\ \text{[H}_2]=\text{[I}_2]=8.03\times 10^{-3} \ M

Check: Plug equilibrium concentrations into the equilibrium expression and calculate K_\text{c}.
\text{[HI]} = 0.242 – 2(8.03 \times 10^{-3}) = 0.226 \ M \\ (8.03 \times 10^{-3})^2/ (0.226)^2= 1.2624 \times 10^{-3}= 1 .26 \times 10^{-3}

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