CHEMICAL EQUILIBRIUM WORKSHEET - Free Printable

Here are the solutions to the problems on your worksheet, broken down step-by-step.

Part 1: Matching Terms

Logic:
* Equilibrium position: This tells us if there is more product or reactant at the end. It depends on where you started (initial concentrations). -> Matches (b).
* Law of chemical equilibrium: This is the general rule that says reactions will always try to reach a balance point ($K_{eq}$). -> Matches (c).
* Reaction quotient ($Q$): This is a calculation used at any moment in time to see if the reaction is balanced yet. If $Q = K$, it's at equilibrium. -> Matches (a).
* Law of mass action: This describes how we calculate the constant using math (products over reactants). -> Matches (d).
* Equilibrium constant ($K_{eq}$): This is the specific number (ratio) you get when the reaction is perfectly balanced. -> Matches (e).

Part 2: Writing Equilibrium Expressions

Rule: The formula is $\frac{\text{Products}}{\text{Reactants}}$.
Important: Do not include Solids $(s)$ or Liquids $(l)$ in the expression. Only Gases $(g)$ and Aqueous solutions $(aq)$ count.

6. $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$
Everything is a gas, so we include everything.
$$K_{eq} = \frac{[HI]^2}{[H_2][I_2]}$$

7. $NH_4Cl(s) \rightleftharpoons NH_3(g) + HCl(g)$
Ignore the solid $NH_4Cl$.
$$K_{eq} = [NH_3][HCl]$$

8. $As_2O_3(s) + 6C(s) \rightleftharpoons As_4(g) + 6CO(g)$
Ignore both solids ($As_2O_3$ and $C$).
$$K_{eq} = [As_4][CO]^6$$

9. $SnO_2(s) + 2CO(g) \rightleftharpoons Sn(s) + 2CO_2(g)$
Ignore the solids ($SnO_2$ and $Sn$).
$$K_{eq} = \frac{[CO_2]^2}{[CO]^2}$$

10. $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$
Ignore both solids. Only the gas remains.
$$K_{eq} = [CO_2]$$

Part 3: Calculating Q and Predicting Direction

Rule: Calculate $Q$ using the current numbers given. Compare $Q$ to $K_{eq}$.
* If $Q < K_{eq}$: Not enough products. Shift Right.
* If $Q > K_{eq}$: Too many products. Shift Left.
* If $Q = K_{eq}$: At Equilibrium. No shift.

11. $2CO(g) \rightleftharpoons C(s) + CO_2(g)$
* $K_{eq} = 7.7 \times 10^{-15}$
* Ignore solid Carbon ($C$). Formula: $Q = \frac{[CO_2]}{[CO]^2}$
* Plug in numbers: $Q = \frac{3.6 \times 10^{-7}}{(0.034)^2} = \frac{3.6 \times 10^{-7}}{0.001156} \approx 3.1 \times 10^{-4}$
* Compare: $3.1 \times 10^{-4}$ is much larger than $7.7 \times 10^{-15}$.
* Answer: Not at equilibrium. Reaction proceeds to the Left.

12. $N_2O_4(g) \rightleftharpoons 2NO_2(g)$
* $K_{eq} = 0.2$
* Formula: $Q = \frac{[NO_2]^2}{[N_2O_4]}$
* Plug in numbers: $Q = \frac{(0.2)^2}{2.0} = \frac{0.04}{2.0} = 0.02$
* Compare: $0.02 < 0.2$ ($Q < K$).
* Answer: Not at equilibrium. Reaction proceeds to the Right.

13. $2ICl(g) \rightleftharpoons I_2(g) + Cl_2(g)$
* $K_{eq} = 0.11$
* Formula: $Q = \frac{[I_2][Cl_2]}{[ICl]^2}$
* Plug in numbers: $Q = \frac{(2.0)(1.2)}{(2.5)^2} = \frac{2.4}{6.25} = 0.384$
* Compare: $0.384 > 0.11$ ($Q > K$).
* Answer: Not at equilibrium. Reaction proceeds to the Left.

14. $Fe_2O_3(s) + 3H_2(g) \rightleftharpoons 2Fe(s) + 3H_2O(g)$
* $K_{eq} = 0.064$
* Ignore solids ($Fe_2O_3$ and $Fe$). Formula: $Q = \frac{[H_2O]^3}{[H_2]^3}$
* Plug in numbers: $Q = \frac{(0.37)^3}{(0.45)^3} = \frac{0.05065}{0.091125} \approx 0.556$
* Compare: $0.556 > 0.064$ ($Q > K$).
* Answer: Reaction proceeds to the Left.

Part 4: Matching Statements about Magnitude

Logic:
* Big $K$ (> 1) means lots of products.
* Small $K$ (< 1) means lots of reactants.
* $K \approx 1$ means a mix of both.

15. Product concentration is much greater than reactants.
This means the top of the fraction is bigger. $K_{eq}$ must be big. -> Matches (a).

16. Product concentration is much less than reactants.
This means the bottom of the fraction is bigger. $K_{eq}$ must be small. -> Matches (c).

17. Considerable amount of both.
This implies a balance, neither side is "winning" completely. -> Matches (b).

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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 & Direction:
11. $Q \approx 3.1 \times 10^{-4}$. Not at equilibrium; shifts Left.
12. $Q = 0.02$. Not at equilibrium; shifts Right.
13. $Q = 0.384$. Not at equilibrium; shifts Left.
14. $Q \approx 0.56$. Shifts Left.

Magnitude Matching:
15. a
16. c
17. b