Answer
Sketch (b) represents an equilibrium mixture.
Work Step by Step
1. Count the atoms in the initial mixture:
4 $SO_2$
2 $Cl_2$
12 $SO_2Cl_2$
2. Now, assume these are the concentrations in the complete mixture, and find the equilibrium amounts.
3. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ SO_2 ]& [ Cl_2 ]& [ SO_2Cl_2 ]\\
Initial& 4 & 2 & 12 \\
Change& -x& -x& +x\\
Equilibrium& 4 -x& 2 -x& 12 +x\\
\end{vmatrix}$$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ SO_2Cl_2 ]}{[ SO_2 ][ Cl_2 ]}$$
4. At equilibrium, these are the concentrations of each compound:
$ [ SO_2 ] = 4 - x$
$ [ Cl_2 ] = 2 - x$
$ [ SO_2Cl_2 ] = 12 + x$
$$4.0 = \frac{(12 + x)}{(4 - x)(2 - x)}$$
5. Solve for x:
$x \approx 1$ or $x \approx 5$
But x cannot be greater than 2, because that would result in a negative Cl2 concentration:
x = 1
So:
$ [ SO_2 ] = 4 - 1= 3$
$ [ Cl_2 ] = 2 - 1= 1$
$ [ SO_2Cl_2 ] = 12 + 1 = 13$
Find the diagram that represents this.
(b)
