Problem 83 - Chemistry

Question

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

Step-by-Step Solution

Answer

The equilibrium constant \(K\) for the given decomposition reaction of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) can be calculated by using the initial mass and volume of the system, the final pressure, and setting up an ICE table. Calculate initial moles, total moles at equilibrium, and changes in moles during the reaction to find equilibrium concentrations. Then, use these equilibrium concentrations to calculate the equilibrium constant \(K\).

Step 1: Find the initial moles

First, we need to find the initial moles of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) using the given mass and the molar mass of the compound. Molar mass of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3} = 5\times 12.01 + 6\times 1.01 + 3\times 16.00 = 114.11 \ \text{g/mol}\). Initial moles \((n_i)\) of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3} = \frac{5.63\ \text{g}}{114.11\ \text{g/mol}}\)

Step 2: Setup the ICE table

Set up an ICE (initial, change, equilibrium) table to record the changes in moles of reactants and products during the reaction. Reaction: $$\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g)$$ | | \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) | \(\mathrm{C}_{2} \mathrm{H}_{6}\) | \(3\ \mathrm{CO}\) | |---------- |------------------------------------ |---------------------------- |-------------- | | Initial | \(n_i\) | 0 | 0 | | Change | \(-x\) | \(+x\) | \(+3x\) | | Equilibrium | \(n_i - x\) | \(x\) | \(3x\) |

Step 3: Calculate final moles using the final pressure and ideal gas law

Using the final pressure \((P_f = 1.63\ \text{atm})\) and ideal gas law, we can find the total moles of gas at equilibrium and use it to calculate \(x\). Ideal Gas Law: \(PV=nRT\) Total moles of gas \((n_t)\) at equilibrium: $$n_t = \frac{PV}{RT} = \frac{(1.63\ \text{atm})(2.50\ \text{L})}{(0.0821\ \text{L.atm/mol.K})(293\ \text{K})}$$ As the total moles of gas at equilibrium are equal to the sum of moles of all the gases, we can write, $$n_t = n_{\text{C}_{5} \text{H}_{6} \text{O}_{3}} + n_{\text{C}_{2} \text{H}_{6}} + n_{\text{CO}} = (n_i - x) + x + 3x$$ Now we can substitute the known values and calculate \(x\).

Step 4: Calculate the equilibrium concentrations

Once we have \(x\), we can find the equilibrium concentrations of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\), \(\mathrm{C}_{2} \mathrm{H}_{6}\), and \(\mathrm{CO}\) using the equilibrium moles and the given volume of the flask. Equilibrium concentration (\(\textit{conc}\)) for each species: \([\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}]_{eq} = \frac{n_{\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}}}{V} = \frac{n_i - x}{2.50\ \text{L}}\) \([\mathrm{C}_{2} \mathrm{H}_{6}]_{eq} = \frac{n_{\mathrm{C}_{2} \mathrm{H}_{6}}}{V} = \frac{x}{2.50\ \text{L}}\) \([\mathrm{CO}]_{eq} = \frac{n_{\mathrm{CO}}}{V} = \frac{3x}{2.50\ \text{L}}\)

Step 5: Calculate the equilibrium constant

Finally, we can calculate the equilibrium constant using the known equilibrium concentrations and the balanced chemical equation: $$K = \frac{[\mathrm{C}_{2} \mathrm{H}_{6}]_{eq} [\mathrm{CO}]_{eq}^3}{[\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}]_{eq}}$$ Now, substitute the values for the equilibrium concentrations and calculate the value of \(K\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction reaches a state where the concentrations of reactants and products remain constant over time. This does not mean the reactions stop entirely but that the forward and reverse reactions occur at the same rate. Therefore, no net change in the amounts of substances occurs.

In the decomposition reaction of \,\( \mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3} \) into \,\( \mathrm{C}_{2} \mathrm{H}_{6} \) and \,\( 3 \mathrm{CO} \), we reach equilibrium when the rate of forming \,\( \mathrm{C}_{2} \mathrm{H}_{6} \) and \,\( \mathrm{CO} \) equals the rate of their recombination to form \,\( \mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3} \).

At equilibrium, the system can be described using the equilibrium constant \( K \), which is a ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients. This value depends on temperature but not on the initial concentrations of reactants and products.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry, expressed as \,\( PV = nRT \). This relation connects the pressure \( P \), volume \( V \), and temperature \( T \) of a gas with the number of moles \( n \) and the ideal gas constant \( R = 0.0821 \ \text{L.atm/mol.K} \).

In the context of the decomposition reaction, this law helps us determine the total number of moles of gas present at equilibrium. By substituting given values of pressure, volume, and temperature of the system, we can calculate the total moles of gases formed.

This calculation is crucial as it allows us to figure out the extent of the reaction and ultimately the value of the equilibrium constant \( K \). Employing the Ideal Gas Law ensures that we understand how gas behaviors influence the equilibrium in gaseous reactions.

Decomposition Reactions

Decomposition reactions involve a single compound breaking down into two or more simpler substances. In the given exercise, the decomposition of \,\( \mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3} \) results in the formation of \,\( \mathrm{C}_{2} \mathrm{H}_{6} \) and \,\( 3 \mathrm{CO} \).

Such reactions often require an external source of heat because breaking the bonds of the original compound requires energy. This specific decomposition reaction is initiated by heating the compound to \( 200. \ ^{\circ} \mathrm{C} \).

Decomposition reactions are significant in various scientific applications, including understanding stability profiles of compounds and determining energy requirements for product formation. This reaction type is fundamental in both academic studies and industrial processes.

ICE Table

The ICE table is a systematic method for keeping track of concentrations or moles of reactants and products in a chemical reaction over time. The acronym ICE stands for Initial, Change, and Equilibrium.

The ICE table begins with initial concentrations or moles before the reaction takes place. Then, based on the stoichiometry of the reaction, change values (usually represented with variables like \( x \)) are used to describe how much the concentrations of reactants decrease and products increase. Finally, the equilibrium row helps us compute their final concentrations or moles.

In the decomposition of \,\( \mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3} \), the ICE table is instrumental to organize how the number of moles of each substance changes and reaches its equilibrium state. This setup allows for straightforward calculations of equilibrium concentrations, enabling the determination of the equilibrium constant \( K \).

By using the ICE table, one can systematically solve complex equilibrium problems with clarity and precision.