Equilibria-kc Question 2

7.4g of SO₂Cl₂ were put in a 2dm³ sealed vessel.

At equilibrium the contained 0.0345mol of Cl₂. Calculate the equilibirum constant for the reaction:

$SO_2Cl_{2(g)}\rightleftharpoons SO_{2(g)}+Cl_{2(g)}$

In this reaction the first thing to note is that the mass and not the moles is given.

It is important to note that the ICE table can only ever have moles, concentration or pressure.

The number of moles of SO₂Cl₂ is:

1 mole SO₂Cl₂ = 135g

7.4g = ?

$\dpi{100} moles \;SO_2Cl_2=\frac{7.4}{135}=0.0548\; moles$

Once the number of moles is noted, then the k_c and the ICE table can be built.

The kc equation is: $\dpi{100} k_c=\frac{[SO_2][Cl_2]}{[SOCl_2]}$

And the ICE table can be written as follows:

Since at equilibrium it is found that the number of moles of SO₂Cl₂ = 0.0345, then:

$0.0548-x = 0.0345$

$\inline \dpi{100} x= \mathbf{{\color{Blue} 0.0203}}$

This would create the number of moles of each component at equilibrium as:

Since the volume is 2dm³, it is important to find the concentration of each component.

$[SO_2Cl_2]=\frac{0.0345}{2}=0.01725M$

$[SO_2]=[Cl_2]=\frac{0.0203}{2}=0.01015M$

Once the concentrations are found, then it is just a matter of placing each value in the k_cequation as follows:

$k_c=\frac{[0.01015][0.01015]}{[0.01725]}$

where k_c would be found to be: 0.597mol dm^-3