Questions: At high temperature, SO2 and NO2 react according to the equation below. Determine the equilibrium constant for this reaction by constructing an ICE table, writing the equilibrium constant expression, and solving it. Complete Parts 1-2 before submitting your answer. SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) A 1.00 L reaction vessel was filled with 2.00 mol SO2 and 2.00 mol NO2. At equilibrium, there were 1.30 mol of NO in the vessel. Fill in the ICE table with the appropriate value for each involved species. SO2(g) NO2(g) SO3(g) + NO(g) Initial (M) Change (M) Equilibrium (M)

Solution Steps

The balanced chemical equation for the reaction is: \[ \mathrm{SO}_{2}(\mathrm{~g}) + \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) + \mathrm{NO}(\mathrm{~g}) \]

Given that the reaction vessel is 1.00 L, the initial concentrations of the reactants are: \[ [\mathrm{SO}_{2}]_0 = 2.00 \, \text{M} \] \[ [\mathrm{NO}_{2}]_0 = 2.00 \, \text{M} \] The initial concentrations of the products are: \[ [\mathrm{SO}_{3}]_0 = 0 \, \text{M} \] \[ [\mathrm{NO}]_0 = 0 \, \text{M} \]

At equilibrium, the concentration of NO is given as 1.30 M. Let \( x \) be the change in concentration for the reactants and products. Since the stoichiometry of the reaction is 1:1:1:1, the change in concentration for each species is: \[ \Delta [\mathrm{NO}] = +1.30 \, \text{M} \] \[ \Delta [\mathrm{SO}_{3}] = +1.30 \, \text{M} \] \[ \Delta [\mathrm{SO}_{2}] = -1.30 \, \text{M} \] \[ \Delta [\mathrm{NO}_{2}] = -1.30 \, \text{M} \]

Using the initial concentrations and the changes, we can find the equilibrium concentrations: \[ [\mathrm{SO}_{2}]_{\text{eq}} = 2.00 - 1.30 = 0.70 \, \text{M} \] \[ [\mathrm{NO}_{2}]_{\text{eq}} = 2.00 - 1.30 = 0.70 \, \text{M} \] \[ [\mathrm{SO}_{3}]_{\text{eq}} = 0 + 1.30 = 1.30 \, \text{M} \] \[ [\mathrm{NO}]_{\text{eq}} = 0 + 1.30 = 1.30 \, \text{M} \]

\[ \begin{array}{|c|c|c|c|c|} \hline & \mathrm{SO}_{2}(\mathrm{~g}) & \mathrm{NO}_{2}(\mathrm{~g}) & \mathrm{SO}_{3}(\mathrm{~g}) & \mathrm{NO}(\mathrm{~g}) \\ \hline \text{Initial (M)} & 2.00 & 2.00 & 0 & 0 \\ \hline \text{Change (M)} & -1.30 & -1.30 & +1.30 & +1.30 \\ \hline \text{Equilibrium (M)} & 0.70 & 0.70 & 1.30 & 1.30 \\ \hline \end{array} \]

The equilibrium constant expression for the reaction is: \[ K_c = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]} \]

Substitute the equilibrium concentrations into the expression: \[ K_c = \frac{(1.30)(1.30)}{(0.70)(0.70)} = \frac{1.69}{0.49} \approx 3.4483 \]

Final Answer