Chemical Equilibrium Worksheet with questions on equilibrium expressions and concepts.

Problem Analysis and Solution

The provided worksheet involves concepts related to chemical equilibrium, including terms like equilibrium position, law of chemical equilibrium, reaction quotient, law of mass action, and equilibrium constant. Additionally, there are questions about writing equilibrium expressions and determining whether a reaction is at equilibrium using the reaction quotient \( Q \).

#### Part 1: Matching Terms
We need to match each term with its correct description.

1. Equilibrium position
- This refers to the state where the concentrations of reactants and products remain constant over time.
- Answer: e. expresses the relative concentration of reactants and products at equilibrium in terms of an equilibrium constant

2. Law of chemical equilibrium
- This states that every reaction proceeds to an equilibrium state with a specific equilibrium constant (\( K_{\text{eq}} \)).
- Answer: c. states that every reaction proceeds to an equilibrium state with a specific \( K_{\text{eq}} \)

3. Reaction quotient
- The reaction quotient (\( Q \)) is used to determine if a reaction has reached equilibrium. It compares the current concentrations of reactants and products to those at equilibrium.
- Answer: a. used to determine if a reaction has reached equilibrium

4. Law of mass action
- This law describes how the equilibrium constant (\( K_{\text{eq}} \)) relates the concentrations of reactants and products at equilibrium.
- Answer: d. expresses the ratio of product concentration to reactant concentration at equilibrium

5. Equilibrium constant
- The equilibrium constant (\( K_{\text{eq}} \)) depends on the initial concentrations of substances in a reaction.
- Answer: b. depends on the initial concentrations of the substances in a reaction

#### Part 2: Writing Equilibrium Expressions
For each given equation, we write the equilibrium expression using the general form:
\[ K_{\text{eq}} = \frac{\text{[Products]}}{\text{[Reactants]}} \]
where solid and liquid phases are not included in the expression.

6. Equation: \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \)
- Equilibrium expression:
\[
K_{\text{eq}} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}
\]

7. Equation: \( \text{NH}_4\text{Cl}(s) \rightleftharpoons \text{NH}_3(g) + \text{HCl}(g) \)
- Since \( \text{NH}_4\text{Cl}(s) \) is a solid, it is not included in the expression.
\[
K_{\text{eq}} = [\text{NH}_3][\text{HCl}]
\]

8. Equation: \( \text{As}_4\text{O}_6(s) + 6\text{C}(s) \rightleftharpoons \text{As}_4(g) + 6\text{CO}(g) \)
- Both \( \text{As}_4\text{O}_6(s) \) and \( \text{C}(s) \) are solids and are not included in the expression.
\[
K_{\text{eq}} = \frac{[\text{As}_4][\text{CO}]^6}{1} = [\text{As}_4][\text{CO}]^6
\]

9. Equation: \( \text{SnO}_2(s) + 2\text{CO}(g) \rightleftharpoons \text{Sn}(s) + 2\text{CO}_2(g) \)
- Both \( \text{SnO}_2(s) \) and \( \text{Sn}(s) \) are solids and are not included in the expression.
\[
K_{\text{eq}} = \frac{[\text{CO}_2]^2}{[\text{CO}]^2}
\]

10. Equation: \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \)
- Both \( \text{CaCO}_3(s) \) and \( \text{CaO}(s) \) are solids and are not included in the expression.
\[
K_{\text{eq}} = [\text{CO}_2]
\]

#### Part 3: Determining Equilibrium Status Using Reaction Quotient (\( Q \))
The reaction quotient \( Q \) is calculated using the same formula as the equilibrium constant but with the current concentrations of reactants and products. If \( Q = K_{\text{eq}} \), the reaction is at equilibrium. If \( Q < K_{\text{eq}} \), the reaction will proceed forward (toward products). If \( Q > K_{\text{eq}} \), the reaction will proceed backward (toward reactants).

11. Reaction: \( 2\text{CO}(g) \rightleftharpoons \text{C}(s) + \text{CO}_2(g) \), \( K_{\text{eq}} = 7.7 \times 10^{-15} \)
- Given: \( [\text{CO}] = 0.034 \, \text{M} \), \( [\text{CO}_2] = 3.6 \times 10^{-17} \, \text{M} \)
- Equilibrium expression:
\[
K_{\text{eq}} = \frac{[\text{CO}_2]}{[\text{CO}]^2}
\]
- Calculate \( Q \):
\[
Q = \frac{[\text{CO}_2]}{[\text{CO}]^2} = \frac{3.6 \times 10^{-17}}{(0.034)^2} = \frac{3.6 \times 10^{-17}}{0.001156} = 3.11 \times 10^{-14}
\]
- Compare \( Q \) and \( K_{\text{eq}} \):
\[
Q = 3.11 \times 10^{-14} \quad \text{and} \quad K_{\text{eq}} = 7.7 \times 10^{-15}
\]
Since \( Q > K_{\text{eq}} \), the reaction will proceed backward (toward reactants).

12. Reaction: \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \), \( K_{\text{eq}} = 0.2 \)
- Given: \( [\text{N}_2\text{O}_4] = 2.0 \, \text{M} \), \( [\text{NO}_2] = 0.2 \, \text{M} \)
- Equilibrium expression:
\[
K_{\text{eq}} = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}
\]
- Calculate \( Q \):
\[
Q = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{(0.2)^2}{2.0} = \frac{0.04}{2.0} = 0.02
\]
- Compare \( Q \) and \( K_{\text{eq}} \):
\[
Q = 0.02 \quad \text{and} \quad K_{\text{eq}} = 0.2
\]
Since \( Q < K_{\text{eq}} \), the reaction will proceed forward (toward products).

13. Reaction: \( 2\text{ICl}(g) \rightleftharpoons \text{I}_2(g) + \text{Cl}_2(g) \), \( K_{\text{eq}} = 0.11 \)
- Given: \( [\text{ICl}] = 2.5 \, \text{M} \), \( [\text{I}_2] = 2.0 \, \text{M} \), \( [\text{Cl}_2] = 1.2 \, \text{M} \)
- Equilibrium expression:
\[
K_{\text{eq}} = \frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2}
\]
- Calculate \( Q \):
\[
Q = \frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2} = \frac{(2.0)(1.2)}{(2.5)^2} = \frac{2.4}{6.25} = 0.384
\]
- Compare \( Q \) and \( K_{\text{eq}} \):
\[
Q = 0.384 \quad \text{and} \quad K_{\text{eq}} = 0.11
\]
Since \( Q > K_{\text{eq}} \), the reaction will proceed backward (toward reactants).

14. Reaction: \( \text{Fe}_2\text{O}_3(s) + 3\text{H}_2(g) \rightleftharpoons 2\text{Fe}(s) + 3\text{H}_2\text{O}(g) \), \( K_{\text{eq}} = 0.064 \) at \( 340^\circ \text{C} \)
- Given: \( [\text{H}_2] = 0.45 \, \text{M} \), \( [\text{H}_2\text{O}] = 0.37 \, \text{M} \)
- Equilibrium expression:
\[
K_{\text{eq}} = \frac{[\text{H}_2\text{O}]^3}{[\text{H}_2]^3}
\]
- Calculate \( Q \):
\[
Q = \frac{[\text{H}_2\text{O}]^3}{[\text{H}_2]^3} = \frac{(0.37)^3}{(0.45)^3} = \frac{0.050653}{0.091125} = 0.556
\]
- Compare \( Q \) and \( K_{\text{eq}} \):
\[
Q = 0.556 \quad \text{and} \quad K_{\text{eq}} = 0.064
\]
Since \( Q > K_{\text{eq}} \), the reaction will proceed backward (toward reactants).

Final Answers

1. Equilibrium position: e
2. Law of chemical equilibrium: c
3. Reaction quotient: a
4. Law of mass action: d
5. Equilibrium constant: b

6. \( K_{\text{eq}} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \)
7. \( K_{\text{eq}} = [\text{NH}_3][\text{HCl}] \)
8. \( K_{\text{eq}} = [\text{As}_4][\text{CO}]^6 \)
9. \( K_{\text{eq}} = \frac{[\text{CO}_2]^2}{[\text{CO}]^2} \)
10. \( K_{\text{eq}} = [\text{CO}_2] \)

11. The reaction will proceed backward.
12. The reaction will proceed forward.
13. The reaction will proceed backward.
14. The reaction will proceed backward.

\boxed{
\begin{array}{ll}
1. & \text{e} \\
2. & \text{c} \\
3. & \text{a} \\
4. & \text{d} \\
5. & \text{b} \\
6. & K_{\text{eq}} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \\
7. & K_{\text{eq}} = [\text{NH}_3][\text{HCl}] \\
8. & K_{\text{eq}} = [\text{As}_4][\text{CO}]^6 \\
9. & K_{\text{eq}} = \frac{[\text{CO}_2]^2}{[\text{CO}]^2} \\
10. & K_{\text{eq}} = [\text{CO}_2] \\
11. & \text{Backward} \\
12. & \text{Forward} \\
13. & \text{Backward} \\
14. & \text{Backward} \\
\end{array}
}