Chemical Equilibrium Worksheet featuring multiple-choice and short-answer questions on equilibrium concepts and calculations.
Chemical Equilibrium Worksheet with questions on equilibrium position, reaction quotient, law of chemical equilibrium, law of mass action, and equilibrium constant, including equations and calculations.
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Step-by-step solution for: CHEMICAL EQUILIBRIUM WORKSHEET
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Show Answer Key & Explanations
Step-by-step solution for: CHEMICAL EQUILIBRIUM WORKSHEET
Here are the solutions to the problems on your worksheet, broken down step-by-step.
Reasoning: We need to match the vocabulary word with its definition based on standard chemistry rules for equilibrium.
1. Equilibrium position: This refers to the specific amounts of reactants and products present when the reaction stops changing. It depends on where you started. (Matches b)
2. Law of chemical equilibrium: This is the fundamental rule stating that at a specific temperature, reactions will always reach a state where the ratio of products to reactants is constant ($K_{eq}$). (Matches c)
3. Reaction quotient ($Q$): This is a calculation used *during* a reaction (before it finishes) to see if it has reached equilibrium yet or which way it needs to shift. (Matches a)
4. Law of mass action: This law describes how to write the expression for the equilibrium constant using concentrations. (Matches d)
5. Equilibrium constant ($K_{eq}$): This is the actual number (the ratio) calculated from the concentrations when the system is at equilibrium. (Matches e)
Rule: The formula is $K_{eq} = \frac{[\text{Products}]}{[\text{Reactants}]}$.
Important Exception: Do not include Solids ($s$) or Liquids ($l$) in the expression. Only Gases ($g$) and Aqueous solutions ($aq$) count.
6. Equation: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$
* All are gases, so all are included.
* Expression: $K_{eq} = \frac{[HI]^2}{[H_2][I_2]}$
7. Equation: $NH_4Cl(s) \rightleftharpoons NH_3(g) + HCl(g)$
* $NH_4Cl$ is a solid ($s$), so we ignore it (treat as 1).
* Expression: $K_{eq} = [NH_3][HCl]$
8. Equation: $As_4O_6(s) + 6C(s) \rightleftharpoons As_4(g) + 6CO(g)$
* Both reactants are solids ($s$), so they are ignored.
* Expression: $K_{eq} = [As_4][CO]^6$
9. Equation: $SnO_2(s) + 2CO(g) \rightleftharpoons Sn(s) + 2CO_2(g)$
* $SnO_2$ and $Sn$ are solids ($s$), so they are ignored.
* Expression: $K_{eq} = \frac{[CO_2]^2}{[CO]^2}$
10. Equation: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$
* Both calcium compounds are solids ($s$), so they are ignored.
* Expression: $K_{eq} = [CO_2]$
Logic:
1. Calculate $Q$ using the same formula as $K_{eq}$.
2. Compare $Q$ to the given $K_{eq}$.
* If $Q < K_{eq}$: Not enough product. Shift Right.
* If $Q > K_{eq}$: Too much product. Shift Left.
* If $Q = K_{eq}$: At Equilibrium.
11. Reaction: $2CO(g) \rightleftharpoons C(s) + CO_2(g)$ | $K_{eq} = 7.7 \times 10^{-15}$
* Ignore Solid C. Formula: $Q = \frac{[CO_2]}{[CO]^2}$
* Math: $Q = \frac{3.6 \times 10^{-7}}{(0.034)^2} = \frac{3.6 \times 10^{-7}}{0.001156} \approx 3.11 \times 10^{-4}$
* Comparison: $3.11 \times 10^{-4}$ is much larger than $7.7 \times 10^{-15}$.
* Answer: No, shift Left.
12. Reaction: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ | $K_{eq} = 0.2$
* Formula: $Q = \frac{[NO_2]^2}{[N_2O_4]}$
* Math: $Q = \frac{(0.2)^2}{2.0} = \frac{0.04}{2.0} = 0.02$
* Comparison: $0.02 < 0.2$.
* Answer: No, shift Right.
13. Reaction: $2ICl(g) \rightleftharpoons I_2(g) + Cl_2(g)$ | $K_{eq} = 0.11$
* Formula: $Q = \frac{[I_2][Cl_2]}{[ICl]^2}$
* Math: $Q = \frac{(2.0)(1.2)}{(2.5)^2} = \frac{2.4}{6.25} = 0.384$
* Comparison: $0.384 > 0.11$.
* Answer: No, shift Left.
14. Reaction: $Fe_2O_3(s) + 3H_2(g) \rightleftharpoons 2Fe(s) + 3H_2O(g)$ | $K_{eq} = 0.064$
* Ignore Solids ($Fe_2O_3, Fe$). Formula: $Q = \frac{[H_2O]^3}{[H_2]^3}$
* Math: $Q = \frac{(0.37)^3}{(0.45)^3} = \frac{0.050653}{0.091125} \approx 0.556$
* Comparison: $0.556 > 0.064$.
* Answer: $Q \approx 0.56$, shift Left.
Logic:
* $K_{eq} > 1$: Products are favored (more products than reactants).
* $K_{eq} < 1$: Reactants are favored (more reactants than products).
* $K_{eq} \approx 1$: Significant amounts of both exist.
15. "The equilibrium concentration of products is much greater..." $\rightarrow$ a ($K_{eq}$ is much greater than 1)
16. "The equilibrium concentration of products is much less..." $\rightarrow$ c ($K_{eq}$ is much less than 1)
17. "There is a considerable amount of both..." $\rightarrow$ b ($K_{eq}$ is about equal to 1)
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Final Answer:
Matching Terms:
1. b
2. c
3. a
4. d
5. e
Equilibrium Expressions:
6. $K_{eq} = \frac{[HI]^2}{[H_2][I_2]}$
7. $K_{eq} = [NH_3][HCl]$
8. $K_{eq} = [As_4][CO]^6$
9. $K_{eq} = \frac{[CO_2]^2}{[CO]^2}$
10. $K_{eq} = [CO_2]$
Calculations:
11. No, shift Left.
12. No, shift Right.
13. No, shift Left.
14. $Q \approx 0.56$, shift Left.
Matching Statements:
15. a
16. c
17. b
Part 1: Matching Terms
Reasoning: We need to match the vocabulary word with its definition based on standard chemistry rules for equilibrium.
1. Equilibrium position: This refers to the specific amounts of reactants and products present when the reaction stops changing. It depends on where you started. (Matches b)
2. Law of chemical equilibrium: This is the fundamental rule stating that at a specific temperature, reactions will always reach a state where the ratio of products to reactants is constant ($K_{eq}$). (Matches c)
3. Reaction quotient ($Q$): This is a calculation used *during* a reaction (before it finishes) to see if it has reached equilibrium yet or which way it needs to shift. (Matches a)
4. Law of mass action: This law describes how to write the expression for the equilibrium constant using concentrations. (Matches d)
5. Equilibrium constant ($K_{eq}$): This is the actual number (the ratio) calculated from the concentrations when the system is at equilibrium. (Matches e)
Part 2: Writing Equilibrium Expressions
Rule: The formula is $K_{eq} = \frac{[\text{Products}]}{[\text{Reactants}]}$.
Important Exception: Do not include Solids ($s$) or Liquids ($l$) in the expression. Only Gases ($g$) and Aqueous solutions ($aq$) count.
6. Equation: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$
* All are gases, so all are included.
* Expression: $K_{eq} = \frac{[HI]^2}{[H_2][I_2]}$
7. Equation: $NH_4Cl(s) \rightleftharpoons NH_3(g) + HCl(g)$
* $NH_4Cl$ is a solid ($s$), so we ignore it (treat as 1).
* Expression: $K_{eq} = [NH_3][HCl]$
8. Equation: $As_4O_6(s) + 6C(s) \rightleftharpoons As_4(g) + 6CO(g)$
* Both reactants are solids ($s$), so they are ignored.
* Expression: $K_{eq} = [As_4][CO]^6$
9. Equation: $SnO_2(s) + 2CO(g) \rightleftharpoons Sn(s) + 2CO_2(g)$
* $SnO_2$ and $Sn$ are solids ($s$), so they are ignored.
* Expression: $K_{eq} = \frac{[CO_2]^2}{[CO]^2}$
10. Equation: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$
* Both calcium compounds are solids ($s$), so they are ignored.
* Expression: $K_{eq} = [CO_2]$
Part 3: Calculating Q and Predicting Direction
Logic:
1. Calculate $Q$ using the same formula as $K_{eq}$.
2. Compare $Q$ to the given $K_{eq}$.
* If $Q < K_{eq}$: Not enough product. Shift Right.
* If $Q > K_{eq}$: Too much product. Shift Left.
* If $Q = K_{eq}$: At Equilibrium.
11. Reaction: $2CO(g) \rightleftharpoons C(s) + CO_2(g)$ | $K_{eq} = 7.7 \times 10^{-15}$
* Ignore Solid C. Formula: $Q = \frac{[CO_2]}{[CO]^2}$
* Math: $Q = \frac{3.6 \times 10^{-7}}{(0.034)^2} = \frac{3.6 \times 10^{-7}}{0.001156} \approx 3.11 \times 10^{-4}$
* Comparison: $3.11 \times 10^{-4}$ is much larger than $7.7 \times 10^{-15}$.
* Answer: No, shift Left.
12. Reaction: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ | $K_{eq} = 0.2$
* Formula: $Q = \frac{[NO_2]^2}{[N_2O_4]}$
* Math: $Q = \frac{(0.2)^2}{2.0} = \frac{0.04}{2.0} = 0.02$
* Comparison: $0.02 < 0.2$.
* Answer: No, shift Right.
13. Reaction: $2ICl(g) \rightleftharpoons I_2(g) + Cl_2(g)$ | $K_{eq} = 0.11$
* Formula: $Q = \frac{[I_2][Cl_2]}{[ICl]^2}$
* Math: $Q = \frac{(2.0)(1.2)}{(2.5)^2} = \frac{2.4}{6.25} = 0.384$
* Comparison: $0.384 > 0.11$.
* Answer: No, shift Left.
14. Reaction: $Fe_2O_3(s) + 3H_2(g) \rightleftharpoons 2Fe(s) + 3H_2O(g)$ | $K_{eq} = 0.064$
* Ignore Solids ($Fe_2O_3, Fe$). Formula: $Q = \frac{[H_2O]^3}{[H_2]^3}$
* Math: $Q = \frac{(0.37)^3}{(0.45)^3} = \frac{0.050653}{0.091125} \approx 0.556$
* Comparison: $0.556 > 0.064$.
* Answer: $Q \approx 0.56$, shift Left.
Part 4: Matching Statements to $K_{eq}$ Values
Logic:
* $K_{eq} > 1$: Products are favored (more products than reactants).
* $K_{eq} < 1$: Reactants are favored (more reactants than products).
* $K_{eq} \approx 1$: Significant amounts of both exist.
15. "The equilibrium concentration of products is much greater..." $\rightarrow$ a ($K_{eq}$ is much greater than 1)
16. "The equilibrium concentration of products is much less..." $\rightarrow$ c ($K_{eq}$ is much less than 1)
17. "There is a considerable amount of both..." $\rightarrow$ b ($K_{eq}$ is about equal to 1)
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Final Answer:
Matching Terms:
1. b
2. c
3. a
4. d
5. e
Equilibrium Expressions:
6. $K_{eq} = \frac{[HI]^2}{[H_2][I_2]}$
7. $K_{eq} = [NH_3][HCl]$
8. $K_{eq} = [As_4][CO]^6$
9. $K_{eq} = \frac{[CO_2]^2}{[CO]^2}$
10. $K_{eq} = [CO_2]$
Calculations:
11. No, shift Left.
12. No, shift Right.
13. No, shift Left.
14. $Q \approx 0.56$, shift Left.
Matching Statements:
15. a
16. c
17. b
Parent Tip: Review the logic above to help your child master the concept of chemical equilibrium worksheet.
