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Chapter 15: Problem 70

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

Expert verified
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

01

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.
02

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.
03

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 \(4.25 \mathrm{g} \ \mathrm{CO}_{2}\)
04

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: \(\frac{0.0966 \ \mathrm{mol}}{10.0 \ \mathrm{L}} = 0.00966 \ \mathrm{M}\)
05

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

\(Q_c = [\mathrm{CO}_2(g)] = 0.00966\)
06

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 \(5.66 \mathrm{g} \ \mathrm{CO}_{2}\)
07

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: \(\frac{0.1286 \ \mathrm{mol}}{10.0 \ \mathrm{L}} = 0.01286 \ \mathrm{M}\)
08

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

\(Q_c = [\mathrm{CO}_2(g)] = 0.01286\)
09

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 \(6.48 \mathrm{g} \ \mathrm{CO}_{2}\)
10

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: \(\frac{0.1473 \ \mathrm{mol}}{10.0 \ \mathrm{L}} = 0.01473 \ \mathrm{M}\)
11

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

\(Q_c = [\mathrm{CO}_2(g)] = 0.01473\)
12

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

At \(373 \mathrm{~K}, K_{p}=0.416\) for the equilibrium \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(\mathrm{~g})\) ?

Ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with halogens \(\left(\mathrm{X}_{2}\right)\) by the following reaction: $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g) $$ The following figures represent the concentrations at equilibrium at the same temperature when \(\mathrm{X}_{2}\) is \(\mathrm{Cl}_{2}\) (green), \(\mathrm{Br}_{2}\) (brown), and \(\mathrm{I}_{2}\) (purple). List the equilibria from smallest to largest equilibrium constant.

Write the equilibrium-constant expression for the equilibrium $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ The table included below shows the relative mole percentages of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{CO}(\mathrm{g})\) at a total pressure of \(1 \mathrm{~atm}\) for several temperatures. Calculate the value of \(K_{p}\) at each temperature. Is the reaction exothermic or endothermic? Explain. $$ \begin{array}{lll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & \mathrm{CO}_{2}(\text { mol } \%) & \text { CO (mol \%) } \\ \hline 850 & 6.23 & 93.77 \\ 950 & 1.32 & 98.68 \\ 1050 & 0.37 & 99.63 \\ 1200 & 0.06 & 99.94 \\ \hline \end{array} $$

The equilibrium constant \(K_{c}\) for \(C(s)+C O_{2}(g) \rightleftharpoons\) \(2 \mathrm{CO}(g)\) is \(1.9\) at \(1000 \mathrm{~K}\) and \(0.133\) at \(298 \mathrm{~K}\). (a) If excess \(\mathrm{C}\) is allowed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) in a 3.00-L vessel at \(1000 \mathrm{~K}\), how many grams of \(\mathrm{CO}_{2}\) are produced? (b) How many grams of \(C\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of \(\mathrm{CO}\) be greater or smaller? (d) If the reaction is endothermic, how does increasing the temperature affect the equilibrium constant?

(a) How is a reaction quotient used to determine whether a system is at equilibrium? (b) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (c) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?
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