Problem 58 - Chemistry The Molecular Nature of Matter

Question

Ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}\), and water react under appropriate conditions to give ethanol. The reaction is: $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) $$ An equilibrium mixture of these gases at a certain temperature had the following partial pressures: \(P_{\mathrm{C}_{2} \mathrm{H}_{4}}=\) $$ 0.575 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=1.30 \mathrm{~atm}, \text { and } P_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}=6.99 \mathrm{~atm} $$ What is the value of \(K_{\mathrm{P}}\) ?

Step-by-Step Solution

Answer

\(K_{\mathrm{P}} = 8.80\) atm.

Step 1: Understanding the Equilibrium Constant (\(K_P\))

The equilibrium constant for a reaction in terms of partial pressures (\(K_P\)) is given by the expression \[K_P = \frac{\text{Partial pressure of products}}{\text{Partial pressure of reactants}}\] Raised to the power of their stoichiometric coefficients in the balanced chemical equation.

Step 2: Write the Expression for the Equilibrium Constant \(K_P\) for the Reaction

For the given reaction \[\mathrm{C}_{2} \mathrm{H}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\] the equilibrium constant expression is \[K_P = \frac{P_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}{P_{\mathrm{C}_{2} \mathrm{H}_{4}} \times P_{\mathrm{H}_{2}\mathrm{O}}}\] because each reactant and product has a stoichiometry of 1 in the balanced reaction.

Step 3: Calculate the Equilibrium Constant \(K_P\)

Given the partial pressures \[P_{\mathrm{C}_2 \mathrm{H}_4} = 0.575\,\text{atm}, P_{\mathrm{H}_2 \mathrm{O}} = 1.30\,\text{atm}, P_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} = 6.99\,\text{atm}\] we substitute these values into the expression for \(K_P\) to get: \[K_P = \frac{6.99}{0.575 \times 1.30}\]

Key Concepts

Chemical Equilibrium

In the study of chemical reactions, the state of chemical equilibrium plays a central role. It occurs when a reaction and its reverse reaction proceed at the same rate. As a result, the amounts of reactants and products remain constant over time. This doesn't mean the quantities are equal, but that their concentrations no longer change. This concept is crucial for understanding how systems react and reach stability under certain conditions.

To gauge the extent of a reaction at equilibrium, we use what is known as the equilibrium constant, denoted as \(K\). It provides a quantitative measure of the proportion of reactants to products. In the case of gases, it is often expressed in terms of partial pressures \(K_P\), whereas with solutions it's usually about concentrations \(K_C\). It's essential to note that \(K\) values are specific to a particular reaction at a particular temperature.

Partial Pressure

The concept of partial pressure is vital when dealing with gaseous systems. The partial pressure of a gas is the pressure it would exert if it alone occupied the entire volume of the mixture at the same temperature. In a mixture of gases, each gas contributes to the total pressure of the mixture. This is known as Dalton's Law of Partial Pressures. Hence, understanding partial pressures allows us to analyze and predict the behavior of gas mixtures in chemical reactions.

In equilibrium problems, one must account for the individual partial pressures of each gas involved to properly calculate the equilibrium constant \(K_P\). For example, if we were to consider a system with multiple gaseous reactants and products, the partial pressures of each would be a determining factor in the value of \(K_P\) and the direction in which the reaction would favor.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is a foundational concept in chemistry that involves ratios derived from the balanced chemical equations. Fundamental stoichiometric calculations allow us to predict the amounts of substances consumed and produced in a given reaction, under the assumption that the reaction goes to completion.

In relation to equilibrium and the equilibrium constant expression, stoichiometry dictates the exponents in the \(K\) expression. For example, if the balanced chemical equation for a reaction has coefficients of 2 for a reactant \(A\) and 1 for a product \(B\), the equilibrium constant expression would be \(K = \frac{[B]}{[A]^2}\), reflecting the stoichiometric ratios. These calculations show how the quantitative relationships are derived and how they dictate the proportions of reactants and products at equilibrium.