Problem 36 - Chemistry The Molecular Science

Question

At high temperature, hydrogen and carbon dioxide react to give water and carbon monoxide. $$ \mathrm{H}_{2 (\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) $$ Laboratory measurements at \(986^{\circ} \mathrm{C}\) show that there is \(0.11 \mathrm{~mol}\) each of \(\mathrm{CO}\) and water vapor and \(0.087 \mathrm{~mol}\) each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) at equilibrium in a sealed 1.0 - \(\mathrm{L}\) container. Calculate the equilibrium constant \(K_{\mathrm{p}}\) for the reac- $$ \text { tion at } 986^{\circ} \mathrm{C} \text { . } $$

Step-by-Step Solution

Answer

The equilibrium constant \( K_p \) at \( 986^{\circ} \mathrm{C} \) is 1.6.

Step 1: Write the Balanced Chemical Equation

For the reaction given, the balanced equation is: \[ \mathrm{H}_{2 (\mathrm{~g})} + \mathrm{CO}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{H}_2 \mathrm{O(\mathrm{g})} + \mathrm{CO(\mathrm{g})} \] This indicates that one mole of hydrogen gas reacts with one mole of carbon dioxide gas to produce one mole of water vapor and one mole of carbon monoxide gas.

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

The equilibrium constant expression for concentration \( K_c \) is given by: \[ K_c = \frac{[\mathrm{H}_2\mathrm{O}][\mathrm{CO}]}{[\mathrm{H}_2][\mathrm{CO}_2]} \]

Step 3: Substitute the Known Concentrations into \( K_c \)

Substitute the values provided for the concentrations into the \( K_c \) expression: \[ [\mathrm{H}_2\mathrm{O}] = 0.11 \, \text{mol/L}, \quad [\mathrm{CO}] = 0.11 \, \text{mol/L} \] \[ [\mathrm{H}_2] = 0.087 \, \text{mol/L}, \quad [\mathrm{CO}_2] = 0.087 \, \text{mol/L} \] Plug these into the expression: \[ K_c = \frac{(0.11)(0.11)}{(0.087)(0.087)} \]

Step 4: Calculate \( K_c \)

Perform the calculation: \[ K_c = \frac{0.0121}{0.007569} \] \[ K_c \approx 1.6 \]

Step 5: Convert \( K_c \) to \( K_p \) Using the Ideal Gas Law

Use the relation between \( K_c \) and \( K_p \): \[ K_p = K_c (RT)^{\Delta n} \] where \( R = 0.0821 \, \text{L atm/K mol} \) and \( \Delta n = (1+1)-(1+1) = 0 \). Since \( \Delta n = 0 \), \( K_p = K_c \). Therefore, \( K_p = 1.6 \).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \) for reactions in terms of concentration, is a vital concept in understanding chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants at equilibrium, each raised to their respective stoichiometric coefficients. For the reaction involving \( \mathrm{H}_2 \) and \( \mathrm{CO}_2 \) forming \( \mathrm{H}_2\mathrm{O} \) and \( \mathrm{CO} \), the balanced chemical equation is critical for writing the correct equilibrium expression. The general form of the equilibrium constant expression for this reaction is:

• \( K_c = \frac{[\mathrm{H}_2\mathrm{O}][\mathrm{CO}]}{[\mathrm{H}_2][\mathrm{CO}_2]} \)

This equation tells us that if the concentration of products is significantly higher than that of reactants, \( K_c \) will be greater than 1, indicating a product-favored reaction. Conversely, if the reactants are more concentrated than the products at equilibrium, \( K_c \) will be less than 1. This expression is derived from the law of mass action, which speaks to the proportional relationships at equilibrium.
It's crucial to note that \( K_c \) depends only on the temperature, hence any change in temperature will alter its value. This makes \( K_c \) a unique constant for a given reaction under specific conditions.

Reaction Quotient

The reaction quotient, \( Q_c \), serves as a snapshot of the reaction's progress at any given moment and is calculated using the same expression as the equilibrium constant:

• \( Q_c = \frac{[\mathrm{H}_2\mathrm{O}][\mathrm{CO}]}{[\mathrm{H}_2][\mathrm{CO}_2]} \)

Yet, unlike \( K_c \), \( Q_c \) can be calculated at any point during the reaction, not just at equilibrium.
This comparison between \( Q_c \) and \( K_c \) provides valuable insight into the status of the chemical reaction:

• If \( Q_c \) = \( K_c \), the system is at equilibrium, indicating no net reaction in the forward or reverse direction.
• If \( Q_c < K_c \), the reaction will proceed in the forward direction to produce more products until equilibrium is achieved.
• If \( Q_c > K_c \), the reaction will shift towards the reactants to reach equilibrium.

By employing the reaction quotient, chemists can predict how a reaction will proceed to reach equilibrium based on the concentrations of reactants and products present at any specific time.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental guideline in predicting how a chemical system at equilibrium responds to disturbances. It states that if an external stress is applied to a system at equilibrium, the system will adjust itself to minimize the effect of that stress, thereby re-establishing equilibrium. This principle can be applied in various scenarios:

• Change in Concentration: Increasing the concentration of reactants will cause the equilibrium to shift toward the products to counteract the change. Conversely, increasing product concentration shifts equilibrium toward reactants.
• Change in Pressure: For reactions involving gases, increasing pressure by decreasing volume will shift equilibrium toward the side with fewer gaseous moles. Reducing pressure will shift it toward more moles.
• Change in Temperature: If the reaction is exothermic (releases heat), increasing temperature will shift the equilibrium toward the reactants. For endothermic reactions, raising the temperature favors product formation.

Understanding Le Chatelier's Principle helps us anticipate how changes influence the position of equilibrium and allows chemists to manipulate conditions to favor desired products or reactants. It's a crucial tool in both industrial applications and theoretical chemistry.