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At $800\:\mathrm{K}$ a reaction mixture contained $0.5\:\mathrm{mol}$ of $\ce{SO2}$, $0.12\:\mathrm{mol}$ of $\ce{O2}$ and $5\:\mathrm{mol}$ of $\ce{SO3}$ at equilibrium. $K_c$ for the equilibrium $$\ce{2SO2 + O2 -> 2SO3}$$ is $833\:\mathrm{L/mol}$. If the volume of the container is $1\:\mathrm{L}$, calculate how much $\ce{O2}$ is to be added at this equilibrium in order to get $5.2\:\mathrm{mol}$ of $\ce{SO3}$ at the same temperature.

The answer comes out to be $0.34\:\mathrm{mol}$.

I have understood the whole question and I am getting the answer. I just have a confusion why we are taking $K_c$ as $833$ and not $1/833$ which makes it $\mathrm{mol/L}$. I think $K_c$ should be in $\mathrm{mol/L}$.

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1 Answer 1

up vote 5 down vote accepted

$K_{c}$ is defined as (for more information see here)

\begin{equation} K_{c} = \prod_i [c_{i}]^{\nu_{i}} = \frac{\prod \limits_{i \in \text{products}} [c_{i}]^{|\nu_{i}|}}{\prod \limits_{j \in \text{reactants}} [c_{j}]^{|\nu_{j}|}} \end{equation}

where $\nu_{i}$ and $c_{i}$ are the stochiometric coefficient and the concentration of the $i^{\text{th}}$ component in the reaction, respectively.

If you use this definition for the reaction

\begin{equation} \ce{2SO2 + O2 -> 2SO3} \end{equation}

you get

\begin{equation} K_{c} = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^{2} [\ce{O2}]} \end{equation}

where the notation $[i] = c_{i}$ was used. This equation tells you that $K_{c}$ has the unit

\begin{equation} \text{unit of }K_{c} = \frac{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2}}{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2} \left(\frac{\mathrm{mol}}{\mathrm{L}} \right)} = \frac{\mathrm{L}}{\mathrm{mol}} \ . \end{equation}

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thanks a lot! Never thought of it with this angle –  user166748 Aug 18 at 17:18

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