Question

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO}\), and \(\mathrm{CO}_{2}\) is placed in a 10.0-L vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO}\), and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO}\), and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (c) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO}\), and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\).

Short answer

For the given mixtures, the amount of \(\mathrm{CaCO}_3\) will change as follows: (a) The amount of \(\mathrm{CaCO}_3\) will decrease, as the reaction shifts to the right. (b) The amount of \(\mathrm{CaCO}_3\) will increase, as the reaction shifts to the left. (c) The amount of \(\mathrm{CaCO}_3\) will increase, as the reaction shifts to the left.

Step-by-step solution

Calculate the initial concentrations of reactants and products

We will use the Ideal Gas Law (\(PV=nRT\)) to calculate the initial concentration of \(\mathrm{CO}_{2}\), as well as converting the masses of reactants and products to moles. It's important to note that the concentrations of both \(\mathrm{CaO}\) and \(\mathrm{CaCO}_{3}\) do not affect the reaction quotient, as they are both solids.

Calculate the reaction quotient \(Q_c\)

It is given by the ratio of the concentrations of products to the concentration of reactants, raised to their respective stoichiometric coefficients. In this case, \(Q_c = [\mathrm{CO}_2(g)]\), as the stoichiometric coefficients for all species are 1 and the solids do not affect the reaction quotient.

Compare \(Q_c\) and \(K_c\)

If \(Q_c > K_c\), the reaction will shift to the left, and if \(Q_c < K_c\), the reaction will shift to the right. If \(Q_c=K_c\), the reaction is already at equilibrium. Now, we will apply the above steps for each given mixture (a), (b), and (c). ## Mixture (a): (a) \(15.0 \mathrm{g} \ \mathrm{CaCO}_{3}\), \(15.0 \mathrm{g} \ \mathrm{CaO}\), and \(4.25 \mathrm{g} \ \mathrm{CO}_{2}\)

Step 1(a): Calculate initial concentrations

Convert grams to moles: Initial moles of \(\mathrm{CO}_{2} = \frac{4.25 \ \mathrm{g}}{44.01 \ \mathrm{g/mol}} = 0.0966 \ \mathrm{mol}\) Initial concentration of CO2: \(\frac{0.0966 \ \mathrm{mol}}{10.0 \ \mathrm{L}} = 0.00966 \ \mathrm{M}\)

Step 2(a): Calculate \(Q_c\)

\(Q_c = [\mathrm{CO}_2(g)] = 0.00966\)

Step 3(a): Compare \(Q_c\) and \(K_c\)

\(Q_c = 0.00966 < K_c = 0.0108\), so the reaction will shift to the right, and the amount of \(\mathrm{CaCO}_{3}\) will decrease. ## Mixture (b): (b) \(2.50 \mathrm{g} \ \mathrm{CaCO}_{3}\), \(25.0 \mathrm{g} \ \mathrm{CaO}\), and \(5.66 \mathrm{g} \ \mathrm{CO}_{2}\)

Step 1(b): Calculate initial concentrations

Convert grams to moles: Initial moles of \(\mathrm{CO}_{2} = \frac{5.66 \ \mathrm{g}}{44.01 \ \mathrm{g/mol}} = 0.1286 \ \mathrm{mol}\) Initial concentration of CO2: \(\frac{0.1286 \ \mathrm{mol}}{10.0 \ \mathrm{L}} = 0.01286 \ \mathrm{M}\)

Step 2(b): Calculate \(Q_c\)

\(Q_c = [\mathrm{CO}_2(g)] = 0.01286\)

Step 3(b): Compare \(Q_c\) and \(K_c\)

\(Q_c = 0.01286 > K_c = 0.0108\), so the reaction will shift to the left, and the amount of \(\mathrm{CaCO}_{3}\) will increase. ## Mixture (c): (c) \(30.5 \mathrm{g} \ \mathrm{CaCO}_{3}\), \(25.5 \mathrm{g} \ \mathrm{CaO}\), and \(6.48 \mathrm{g} \ \mathrm{CO}_{2}\)

Step 1(c): Calculate initial concentrations

Convert grams to moles: Initial moles of \(\mathrm{CO}_{2} = \frac{6.48 \ \mathrm{g}}{44.01 \ \mathrm{g/mol}} = 0.1473 \ \mathrm{mol}\) Initial concentration of CO2: \(\frac{0.1473 \ \mathrm{mol}}{10.0 \ \mathrm{L}} = 0.01473 \ \mathrm{M}\)

Step 2(c): Calculate \(Q_c\)

\(Q_c = [\mathrm{CO}_2(g)] = 0.01473\)

Step 3(c): Compare \(Q_c\) and \(K_c\)

\(Q_c = 0.01473 > K_c = 0.0108\), so the reaction will shift to the left, and the amount of \(\mathrm{CaCO}_{3}\) will increase.

Key concepts

Reaction Quotient

The reaction quotient, expressed as \( Q \), is a crucial concept in understanding chemical equilibrium. It is a snapshot of a reaction's current state compared to its equilibrium state. For the reaction \( \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \), the reaction quotient, \( Q_c \), is determined solely by the concentration of \( \mathrm{CO}_{2}(g) \) since \( \mathrm{CaCO}_{3} \) and \( \mathrm{CaO} \) are solids and do not appear in the expression. When calculating \( Q_c \), use the formula \( Q_c = [\mathrm{CO}_2(g)] \). This means that \( Q_c \) relies on the concentration of the gas, providing a simple way to track the reaction's progression towards equilibrium.

By comparing \( Q_c \) to the equilibrium constant \( K_c \), chemists can predict whether a reaction will shift to the right (products) or left (reactants) to reach equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept to predict how a system at equilibrium will respond to changes in concentration, pressure, or temperature. It states that if an external change is applied to a system at equilibrium, the system will adjust itself to partially counteract the effect of the change.

In the context of the given reaction, if \( Q_c < K_c \), the principle predicts that the reaction will proceed in the forward direction, consuming reactants and producing more products until equilibrium is restored. Conversely, if \( Q_c > K_c \), the reaction will shift in the reverse direction to form more reactants until \( Q_c \) equals \( K_c \). This understanding allows chemists to manipulate reactions to increase yields or reduce by-products by adjusting conditions such as concentrations or pressures.

Equilibrium Constant

An equilibrium constant, designated as \( K \), quantifies the ratio of concentrations of products to reactants for a reaction at equilibrium. At a given temperature, \( K \) measures the balance between reactants and products.

• For our reaction \( \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \), the equilibrium constant \( K_c \) is given as 0.0108 at \( 900^{\circ}\text{C} \).

• \( K_c \) tells us the relative concentrations of products and reactants at equilibrium, providing insights into the reaction's extent; a large \( K_c \) suggests more products than reactants, indicating the reaction favors products.

When \( Q_c \) is compared to \( K_c \), it becomes evident whether the current state of the system is at equilibrium. The equilibrium constant remains unchanged unless there is a temperature change, making it a reliable metric for predicted equilibrium shifts and outcomes under given conditions.