Equilibrium15 Markswrite The Expression For The Equilibrium Const ✓ Solved

EQUILIBRIUM: (1.5 marks) Write the expression for the equilibrium constants for the following reactions: a. N2(g) + 3H2(g) 2NH3(g) b. 2SO 2(g) + O2(g) 2SO3(g) c. NH 3(g) + HCl(g) NH4Cl(s) (3 marks – 1 here, 2 on next page) For reaction 2(a) above, suppose equilibrium has been established. Explain how and why the position of equilibrium will shift in response to the following changes: a.

Additional nitrogen is added to the system. b. Ammonia is removed from the system as it forms. c. The pressure on the system is increased. (1.5 marks) A 1.00 mL sample of dilute acetic acid required 23.4 mL of 0.10 M sodium hydroxide solution to titrate the sample to a phenolphthalein endpoint (which is assumed to be close in pH to the equivalence point). What was the concentration of acetic acid in the sample (show your work)? How many significant figures are justified in your answer? Why?

Paper for above instructions

Equilibrium and Reaction Constants

Equilibrium in chemical reactions is a state where the reactants and products are present at concentrations that do not change with time, indicating that the forward and reverse reactions happen at equal rates. We can express the equilibrium constant (\( K \)) for any reversible reaction in terms of the concentrations of the products and reactants at equilibrium.

1. Equilibrium Constant Expressions for Given Reactions

a. Reaction:
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
Equilibrium Constant Expression:
\[
K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}
\]
In this expression, \( [NH_3] \) is the concentration of ammonia, while \( [N_2] \) and \( [H_2] \) are the concentrations of nitrogen and hydrogen, respectively.
---
b. Reaction:
\[
2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)
\]
Equilibrium Constant Expression:
\[
K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}
\]
For this equilibrium, \( [SO_2] \) and \( [O_2] \) are the concentrations of sulfur dioxide and oxygen, while \( [SO_3] \) refers to sulfur trioxide.
---
c. Reaction:
\[
NH_3(g) + HCl(g) \rightleftharpoons NH_4Cl(s)
\]
Equilibrium Constant Expression:
\[
K_c = \frac{1}{[NH_3][HCl]}
\]
In this case, because ammonium chloride (\( NH_4Cl \)) is a solid, it does not appear in the expression for \( K_c \).

2. Position of Equilibrium in Reaction 2(a)

For the reaction
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
the position of the equilibrium can shift in response to various changes according to Le Chatelier's Principle, which states that when a system at equilibrium is disturbed, it will shift in a direction that counteracts the disturbance.
a. Effect of Adding Additional Nitrogen
When additional \( N_2 \) is added to the system, the concentration of \( N_2 \) increases. According to Le Chatelier’s Principle, the equilibrium will shift to the right to consume the excess \( N_2 \), favoring the formation of ammonia \( NH_3 \) until a new equilibrium is established.
b. Effect of Removing Ammonia
If ammonia \( NH_3 \) is continuously removed from the system as it forms, the concentration of \( NH_3 \) decreases. To counteract this change, the equilibrium will shift to the right to produce more \( NH_3 \) from nitrogen and hydrogen, thereby replenishing the concentration of ammonia until equilibrium is again achieved.
c. Effect of Increasing Pressure
In reactions involving gases, increasing the pressure shifts the equilibrium towards the side with fewer moles of gas. In this reaction, 1 mole of \( N_2 \) and 3 moles of \( H_2 \) (totaling 4 moles of reactants) are transformed into 2 moles of \( NH_3 \) (products). Therefore, increasing the pressure will shift the equilibrium to the right, favoring the production of ammonia because the right side has fewer moles of gas.

3. Concentration of Acetic Acid

To determine the concentration of acetic acid in a 1.00 mL sample that required 23.4 mL of 0.10 M sodium hydroxide (NaOH) for titration, we can use the relationship between volume and concentration given by the equation:
\[
C_1V_1 = C_2V_2
\]
Where:
- \( C_1 \) is the concentration of acetic acid
- \( V_1 \) is the volume of acetic acid (1.00 mL)
- \( C_2 \) is the concentration of NaOH (0.10 M)
- \( V_2 \) is the volume of NaOH used (23.4 mL)
Since the reaction between acetic acid (\( CH_3COOH \)) and NaOH is a 1:1 molar reaction:
\[
C_{acetic \ acid} \times 1.00 \ \text{mL} = 0.1 \ \text{M} \times 23.4 \ \text{mL}
\]
Calculating \( C_{acetic \ acid} \):
\[
C_{acetic \ acid} = \frac{0.1 \times 23.4}{1.00} = 2.34 \ \text{M}
\]

Significant Figures

The number of significant figures in the final answer (2.34 M) can be determined based on the values provided in the calculations. The measurement for \( NaOH \) was given as 23.4 mL, which has three significant figures, and the concentration of NaOH is also expressed with two significant figures. However, since the final result is determined by the volume used (23.4 mL), the final answer should therefore be expressed with three significant figures, resulting in:
Concentration of Acetic Acid: \( 2.34 \ \text{M} \)

Conclusion

The concept of equilibrium and the calculations involving equilibrium constants are fundamental in understanding chemical reactions. The adjustments in position due to external changes, such as alterations in concentration, pressure, or temperature, significantly impact the amounts of reactants and products at equilibrium. This assessment reinforces the critical relationship between concentration, volume, and the principles governing chemical equilibria.

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