Question

Consider the decomposition of the compound \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) as follows: $$ \mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g) $$ When a 5.63-g sample of pure \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g)\) was sealed into an otherwise empty 2.50-L flask and heated to \(200 .{ }^{\circ} \mathrm{C}\), the pressure in the flask gradually rose to \(1.63\) atm and remained at that value. Calculate \(K\) for this reaction.

Short answer

The equilibrium constant, \(K\), for the decomposition reaction of the compound \(\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}(g)\) at \(200 .{ }^{\circ}\mathrm{C}\) is approximately 57.8.

Step-by-step solution

Calculate the number of moles of \(\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}(g)\)

First, we need to find the molar mass of the compound: \(\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3} = 5(12.01)+6(1.01)+3(16.00) \approx 126.08\ \text{g/mol}\) Now, we can find the moles of \(\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}\) using its mass: \(moles_{\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}} = \frac{5.63\ \text{g}}{126.08\ \text{g/mol}} \approx 0.0446\ \text{mol}\)

Create an ICE table

Let's set up an ICE table to represent the stoichiometry of the reaction: | | \(\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}\) | \(\mathrm{C}_{2}\mathrm{H}_{6}\) | \(\mathrm{CO}\) | |:-------------:|:-------------------:|:----------:|:-----:| | Initial (mol) | 0.0446 | 0 | 0 | | Change (mol) | -x | x | 3x | | Eq. (mol) | 0.0446 - x | x | 3x | At equilibrium, the number of moles of \(\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}\) and \(\mathrm{C}_{2}\mathrm{H}_{6}\) change by -x and x, respectively, while the number of moles of \(\mathrm{CO}\) changes by 3x.

Determine equilibrium concentrations

We are given that the final pressure at equilibrium is 1.63 atm. Using the ideal gas law, we can find the total number of moles of gas at equilibrium: $$ PV = nRT $$ $$ n_{total} = \frac{PV}{RT} = \frac{(1.63\ \text{atm})(2.50\ \text{L})}{(0.08206\ \text{L\ atm/mol K}) [ 200 .{ }^{\circ}\mathrm{C} + 273.15\ \text{K} ]} \approx 0.1482\ \text{mol} $$ From the ICE table, we know that at equilibrium, the total number of moles is \(0.0446 - x + x + 3x = 0.0446 + 2x\). Therefore: $$ 0.0446 + 2x \approx 0.1482\ \text{mol} \\ 2x \approx 0.1036 \\ x \approx 0.0518 $$ Now we can determine the concentrations of the reactant and products at equilibrium by dividing their moles by the volume of the flask (2.50 L): - \(\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}(g)\): \(\frac{0.0446 - 0.0518}{2.50}\ \text{L} \approx 0.00944\ \text{M}\) - \(\mathrm{C}_{2}\mathrm{H}_{6}(g)\): \(\frac{0.0518}{2.50}\ \text{L} \approx 0.0207\ \text{M}\) - \(\mathrm{CO}(g)\): \(\frac{3(0.0518)}{2.50}\ \text{L} \approx 0.0621\ \text{M}\)

Calculate the equilibrium constant, \(K\)

Now that we have the equilibrium concentrations, we can calculate the equilibrium constant, \(K\), using the following expression: $$ K = \frac{[C_2H_6][CO]^3}{[C_5H_6O_3]} $$ $$ K = \frac{(0.0207)(0.0621)^3}{(0.00944)} \approx 57.8 $$ Thus, the equilibrium constant, \(K\), for the given decomposition reaction is approximately 57.8.

Key concepts

Understanding Chemical Equilibrium Concentrations

When studying chemical equilibrium, it's essential to comprehend how the concentrations of substances in a system change and settle into a steady-state. Equilibrium concentrations refer to the amounts of reactants and products in a chemical reaction that remain constant over time at a given temperature and pressure, assuming the system is closed.

During a reaction, as reactants are consumed, products are formed until a point is reached where the rate of the forward reaction equals that of the reverse reaction. At this point, the reaction does not stop but continues with both the forward and reverse reactions occurring at the same rate, leading to constant concentrations of the reactants and products. These concentrations are what we refer to when we talk about Equilibrium Concentrations.

Calculating these concentrations can be complex, requiring a deep understanding of reaction stoichiometry and the conditions of the system, such as temperature and pressure. One common method for determining these concentrations is the Initial, Change, Equilibrium (ICE) table method, which helps track the change in concentration of reactants and products over time in a systematic way.

The ICE Table Method Simplified

The ICE Table Method is a practical tool for visualizing and calculating the changes in concentration from the beginning of a reaction up to equilibrium. ICE stands for Initial, Change, and Equilibrium - representing different stages of the reaction.

Initial:

This represents the initial concentrations of the reactants and products before the reaction starts. Products are usually zero for reactions starting from reactants only.

Change:

It shows the change in concentrations of reactants and products as the reaction proceeds. These changes are expressed in terms of a variable, often 'x', representing the amount of reactant that reacts.

Equilibrium:

We calculate these values by combining the initial concentrations with the changes that have occurred. The equilibrium concentrations are then used to calculate the equilibrium constant.

This method is extremely useful for both learning and problem-solving because it provides a clear and organized way to approach equilibrium calculations, breaking them down into manageable steps.

Applying the Ideal Gas Law to Equilibrium

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and number of moles of a gas to each other. It is expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

For reactions involving gases, the Ideal Gas Law becomes crucial for determining the equilibrium concentrations. This is because, at equilibrium, we can measure the total pressure exerted by the gaseous mixture, and by knowing the temperature and volume of the container, we can use the law to calculate the total number of moles of gas present.

Once we have the total moles and the equilibrium moles of individual components (as determined by the ICE Table), we can easily find the molar concentration of each gas by dividing its moles by the volume of the system (in liters). This step is vital to calculate the equilibrium constant (K), which provides an invaluable insight into the extent of the reaction under given conditions and helps predict the behavior of the reaction in response to changes in conditions, based on Le Chatelier's principle.