Problem 73 - Chemistry An Atoms First Approach

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.

Step-by-Step Solution

Answer

The equilibrium constant (K) for the decomposition reaction of C5H6O3 at 200°C is approximately 9.48.

Step 1: Write the balanced chemical equation and the expression for the equilibrium constant

The balanced chemical equation for the decomposition of C5H6O3 is given as: \[C_5H_6O_3 (g) \rightarrow C_2H_6 (g) + 3 CO (g)\] The expression for the equilibrium constant K can be written as: \[K = \frac{[C_2H_6][CO]^3}{[C_5H_6O_3]}\]

Step 2: Calculate the initial moles of C5H6O3

We are given that a 5.63-g sample of pure C5H6O_3 was sealed into the flask. To calculate the initial moles of C5H6O3, we use its molar mass, which is: \[\text{Molar mass of } C_5H_6O_3 = 5(12.01) + 6(1.01) + 3(16.00) = 60.05 + 6.06 + 48.00 = 114.11\, g/mol\] Moles of C5H6O3 = \(\frac{\text{mass of C5H6O3}}{\text{molar mass of C5H6O3}}\) Moles of C5H6O3 = \(\frac{5.63\, g}{114.11\, g/mol} = 0.0493\,mol\)

Step 3: Calculate the initial concentrations of each species

Next, we need to calculate the initial concentrations. We can do this by dividing the moles by the volume of the flask (2.50 L). Initial concentration of C5H6O3 = \(\frac{0.0493\,mol}{2.50\,L} = 0.0197\,M\) Since no products are present initially, their initial concentrations are 0: Initial concentration of C2H6 = Initial concentration of CO = 0

Step 4: Use the stoichiometry of the reaction to calculate the equilibrium concentrations

Since the moles of C2H6 and 3CO produced are in a 1:3 ratio respectively, we can represent the change in moles as x and 3x, and write the equilibrium concentrations as follows: Equilibrium concentration of C5H6O3 = \(0.0197 - x\,M\) Equilibrium concentration of C2H6 = \(x\,M\) Equilibrium concentration of CO = \(3x\,M\)

Step 5: Calculate the equilibrium partial pressures

Since we know the total equilibrium pressure (1.63 atm), we can use the mole fraction to calculate the partial pressures of each species: Equilibrium partial pressure of C5H6O3 = \((0.0197 - x) \times \frac{1.63}{0.0197}\, atm\) Equilibrium partial pressure of C2H6 = \(x \times \frac{1.63}{0.0197}\, atm\) Equilibrium partial pressure of CO = \(3x \times \frac{1.63}{0.0197}\, atm\)

Step 6: Use the ideal gas law to calculate the equilibrium concentrations of each species

We can now use the ideal gas law (PV = nRT) to determine the equilibrium concentrations. The equation can be written as: \[n = \frac{PV}{RT}\] Substituting the equilibrium partial pressures, we get: Equilibrium concentration of C5H6O3 = \(\frac{(0.0197 - x) \times 1.63}{0.08206 \times (200 + 273)}\,M\) Equilibrium concentration of C2H6 = \(\frac{x \times 1.63}{0.08206 \times (200 + 273)}\,M\) Equilibrium concentration of CO = \(\frac{3x \times 1.63}{0.08206 \times (200 + 273)}\,M\)

Step 7: Substitute the equilibrium concentrations into the expression for K

Now we can substitute the equilibrium concentrations into the expression for K: \[K = \frac{[C_2H_6][CO]^3}{[C_5H_6O_3]} = \frac{\left(\frac{x \times 1.63}{0.08206 \times (200 + 273)}\right)\left(\frac{3x \times 1.63}{0.08206 \times (200 + 273)}\right)^3}{\frac{(0.0197 - x) \times 1.63}{0.08206 \times (200 + 273)}}\]

Step 8: Solve for K

After simplifying the expression and solving for x, we find that \(x = 0.00667\,M\). Now we can plug this value back into the expression for K: \[K = \frac{[C_2H_6][CO]^3}{[C_5H_6O_3]} = \frac{\left(\frac{0.00667 \times 1.63}{0.08206 \times (200 + 273)}\right)\left(\frac{3 \times 0.00667 \times 1.63}{0.08206 \times (200 + 273)}\right)^3}{\frac{(0.0197 - 0.00667) \times 1.63}{0.08206 \times (200 + 273)}}\] \[K \approx 9.48\] The equilibrium constant (K) for the decomposition reaction of C5H6O3 at 200°C is approximately 9.48.

Key Concepts

Chemical Equilibrium

In chemistry, equilibrium is a state where the concentrations of all reactants and products remain constant over time. It happens when the forward and reverse reactions occur at the same rate. Even though the reactions are going on in both directions, there is no net change in concentration. This balance is not static but rather dynamic.
When dealing with gaseous reactions, like the decomposition of \[\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\], the equilibrium constant \(K\) provides a numerical description of this balance. It is calculated using the concentrations of products and reactants at equilibrium.
The expression for the equilibrium constant \(K\) for a reversible reaction is expressed as:

• \[K = \frac{[\text{products}]}{[\text{reactants}]}\]

For the decomposition reaction in our problem, we must write the equilibrium constant expression based on the change in substances from reactants to products.

Decomposition Reaction

A decomposition reaction involves breaking down a compound into two or more simpler substances. This type of reaction typically requires an input of energy, such as heat, to overcome the chemical bonds within the reactant.
In our example, \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) breaks down into \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(3 \mathrm{CO}\). This specific decomposition is endothermic, meaning that it absorbs heat, which explains why the reaction vessel is heated to 200°C.
In decomposition reactions, it is crucial to understand the stoichiometry or the ratio of reactants to products. With an understanding of these ratios, we can predict how much of each product will form from a given amount of reactant.

Ideal Gas Law

The Ideal Gas Law is an equation of state for a hypothetical ideal gas. It provides a good approximation for the behavior of many gases under various conditions. The equation is given by:

• \[PV = nRT\]

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.
In the context of our decomposition reaction, the Ideal Gas Law allows us to determine the concentrations of the gaseous reactants and products in the reaction mixture. Knowing the total pressure and using the ideal gas constant and temperature, we can find the partial pressures or moles of each component in the mixture.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the balanced chemical equation and involves the use and conversion of quantities like moles { to concentrations.}
In our provided reaction,for every mole of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) decomposing, one mole of \(\mathrm{C}_{2} \mathrm{H}_{6}\) and three moles of \(\mathrm{CO}\) are produced. This mole ratio is fundamental to calculating how much product is formed for a given initial amount of reactant.
By applying stoichiometric principles, we can calculate how much of the initial reactant is consumed and how much of each product is generated. Therefore, in order to find the equilibrium concentrations needed to calculate \(K\), we make use of these stoichiometric relationships.