Question

At a particular temperature, 12.0 moles of SO₃ is placed into a 3.0-L rigid container, and the SO₃ dissociates by the reaction

${\mathbf{\text{2SO}}}_{\mathbf{\text{3}}}\mathbf{\text{(g)}}\mathbf{\rightleftharpoons }{\mathbf{\text{2SO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) +O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

At equilibrium, 3.0 moles of SO₂ is present. Calculate K for this reaction.

Short answer

The K value for this reaction at this temperature is 0.056.

Step-by-step solution

Step1:

The expression for the equilibrium constant, K of this reaction in terms of concentrations is written below:

$\text{K =}\frac{{\left[{\text{SO}}_{\text{2}}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]}{{\left[{\text{SO}}_{\text{3}}\right]}^{\text{2}}}$ …(1)

The number of moles ofis given as 12.0 mol.

The volume of the container is given as 3.0 L.

Hence, the concentration of${\text{SO}}_{\text{3}}$is,

$\begin{array}{rcl}& & \left[{\text{SO}}_{\text{3}}\right]\text{=}\frac{\text{Moles} \text{of} {\text{SO}}_{\text{3}}}{\text{Volume}}\\ & & \text{=}\frac{\text{12.0} \text{mol}}{\text{3.0} \text{L}}\\ & & \text{= 4.0} \text{M}\\ & & \end{array}$

 Step 2:

Let x be the concentration of${\text{SO}}_{\text{3}}$consumed to reach the equilibrium.

Setup the ICE table:

$$\begin{gathered} {\text{2SO}}_{\text{3}}\left(\text{g}\right){\text{⇔2SO}}_{\text{2}}\left(\text{g}\right){\text{+O}}_{\text{2}}\left(\text{g}\right) \\ \text{Iinitial}\left(\text{M}\right)\text{4.000} \\ \text{Change}\left(\text{M}\right) \text{-} 2\text{x} \text{-} 2\text{x+ x} \\ \text{Equilibrium}\left(\text{M}\right)\text{4.0} \text{-} 2\text{x}2\text{xx} \end{gathered}$$

Substitute these equilibrium concentrations in terms of x into equation (1).

localid="1657791912320" $\text{K =}\frac{{\left(\text{2x}\right)}^{\text{2}}\left(\text{x}\right)}{{\left(\text{4.0} \text{-} 2\text{x}\right)}^{\text{2}}}$ …(2)

 Step 3:

The number of moles of${\text{SO}}_{\text{2}}$present at equilibrium is given as 3.0 mol. Hence, the concentration of at equilibrium is,

$\begin{array}{rcl}& & \left[{\text{SO}}_{\text{2}}\right]\text{=}\frac{\text{Moles} \text{of} {\text{SO}}_{\text{2}}}{\text{Volume}}\\ & & \text{=}\frac{\text{3.0} \text{mol}}{\text{3.0} \text{L}}\\ & & \text{= 1.0} \text{M}\\ & & \end{array}$

From the ICE table, the equilibriumconcentration of is 2x. Hence,

$$\begin{gathered} \text{2x = 1.0} \\ \text{x = 0.5} \end{gathered}$$

Therefore, the x value is 0.5.

Step 4:

Substitute this value into Equation (2) to calculate the K as shown below:

$\begin{array}{rcl}& & \text{K =}\frac{{\left(2\times \text{0.5}\right)}^{\text{2}}\left(\text{0.5}\right)}{{\left(\text{4.0} \text{-} \text{2×0.5}\right)}^{\text{2}}}\\ & & \\ & & \text{=}\frac{{\left(\text{1.0}\right)}^{\text{2}}\left(\text{0.5}\right)}{{\left(\text{3.0}\right)}^{\text{2}}}\\ & & \text{= 0.056}\\ & & \end{array}$

Hence, the K value for this reaction at this temperature is 0.056.