Question

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a 2.00 -L flask. At \(648 \mathrm{~K}\), there is 0.0345 mole of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$

Short answer

\(K_{c} = 0.0192\)

Step-by-step solution

Mole Calculations

Calculate the number of moles of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\). Using the molar mass of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) which is \(102.97 \mathrm{~g}/\mathrm{mol}\), the number of moles can be worked out as \(6.75 \mathrm{~g} / 102.97 \mathrm{~g}/\mathrm{mol} = 0.0656 \mathrm{~mol}\)

Determine the Initial and Equilibrium Amounts

At the start of the reaction, there are \(0.0656 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) and 0 of \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\). At equilibrium, \(0.0345 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\) is present. Hence the same quantity of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) must have decomposed and the same amount of \(\mathrm{Cl}_{2}\) must have formed. So the mols at equilibrium are: \(\mathrm{SO}_{2}\mathrm{Cl}_{2} = 0.0656 \mathrm{~mol}-0.0345 \mathrm{~mol} = 0.0311 \mathrm{~mol}\), \(\mathrm{SO}_{2} = 0.0345 \mathrm{~mol}\), and \(\mathrm{Cl}_{2} = 0.0345 \mathrm{~mol}\)

Find the Equilibrium Concentrations

To convert the number of moles to concentration (in M), divide by the volume of the flask: \([\mathrm{SO}_{2}\mathrm{Cl}_{2}] = 0.0311 \mathrm{~mol} / 2.00 \mathrm{~L} = 0.01555 \mathrm{~M}\), \([\mathrm{SO}_{2}] = 0.0345 \mathrm{~mol} / 2.00 \mathrm{~L} = 0.01725 \mathrm{~M}\) and \([\mathrm{Cl}_{2}] = 0.0345 \mathrm{~mol} / 2.00 \mathrm{~L} = 0.01725 \mathrm{~M}\)

Calculate \(K_{c}\)

The equilibrium constant \(K_{c}\) for the reaction is given by the expression: \(K_{c} = [\mathrm{SO}_{2}] [\mathrm{Cl}_{2}] / [\mathrm{SO}_{2}\mathrm{Cl}_{2}] = (0.01725 \times 0.01725) / 0.01555 = 0.0192\)

Key concepts

Mole Calculations

Mole calculations provide us with a way to convert between the mass of a substance and the moles, which are a fundamental unit in chemistry for counting particles. This is crucial for understanding chemical reactions. In our example, we started with

• Mass: 6.75 g of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \)
• Molar Mass: 102.97 g/mol of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \)

Using these values, you can calculate the moles of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) by dividing the given mass by the molar mass: 6.75 g / 102.97 g/mol = 0.0656 mol.
This tells us how many moles of molecules we initially had, which is important for determining the progress of the chemical reaction. Mole calculations like this are foundational, as they allow us to work with chemical formulas and amounts accurately.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of reactants and products in a chemical reaction that remain constant over time due to an established state of equilibrium. For any chemical reaction occurring in a closed system, the point at which the rate of the forward reaction equals the rate of the backward reaction is called equilibrium.
In the example given, when the reaction reaches equilibrium at 648 K, the concentrations are:

• \([\mathrm{SO}_{2}\mathrm{Cl}_{2}] = 0.0311\, \mathrm{mol} / 2.00 \mathrm{~L} = 0.01555 \mathrm{~M}\)
• \([\mathrm{SO}_{2}] = 0.0345\, \mathrm{mol} / 2.00 \mathrm{~L} = 0.01725 \mathrm{~M}\)
• \([\mathrm{Cl}_{2}] = 0.0345\, \mathrm{mol} / 2.00 \mathrm{~L} = 0.01725 \mathrm{~M}\)

These concentrations allow us to calculate the equilibrium constant \( K_c \), which gives us insight into the relative amounts of reactants and products at equilibrium, thus quantitatively describing the position of the equilibrium.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. This process involves breaking and forming bonds, leading to a rearrangement of atoms. The given reaction where \(\mathrm{SO}_{2}\mathrm{Cl}_{2} \) decomposes into \(\mathrm{SO}_{2} \) and \(\mathrm{Cl}_{2} \) is an example of a reversible reaction. In a closed system, such reactions can reach a state of dynamic equilibrium.
When writing a chemical equation, the reactants are shown on the left and products on the right with an arrow between them. The double-sided arrow in this specific reaction, \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\), indicates that the reaction can proceed in both directions.
The forward reaction forms products, while the backward reaction forms reactants, and over time, they balance out, leading to the equilibrium state. Understanding these principles is essential for predicting how changes in conditions might affect the equilibrium state, a central tenet of chemical dynamics.