Unit 8: Chemical Equilibrium — Short Questions

Ans. It is the state in a reversible direction when the rates of forward and reverse reactions become equal and the concentrations of reactants and products remain constant.

Ans. A reversible reaction is one that proceeds in both forward and reverse directions.

Example
N₂ + 3H₂(g) ⇌ 2NH₃(g)

Ans. Volume change and equilibrium:
Change in volume affects the equilibrium position, not the equilibrium constant, which only changes with temperature.

Ans. Characteristics of chemical equilibrium:

• Forward and reverse reaction rates are equal
• Concentrations remain constant
• Can be reached from either side
• Dynamic in nature

Ans. Because reactants and products continuously convert into each other at equal rates; the system is active even though no visible change occurs. It is macroscopically dynamic in nature.

Ans. Because the concentration of reactants decreases, lowering the rate of the forward reaction until it equals the reverse reaction.

Ans. Because ice has a lower density than water. Applying pressure lowers the melting point, causing it to melt even without heat.

Ans. The conditions of equilibrium constant are

• Temperature must remain constant.
• It applies only to a system in dynamic equilibrium.

Ans.
We are given the reversible reaction
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Initial concentrations

• [SO₂] = 0.60 mol/dm³
• [O₂] = 0.50 mol/dm³
• [SO₃] = 0 initially
• At equilibrium [SO₃] = 0.040 mol/dm³

Step 1 Use stoichiometry to find change in concentration.
Since the balanced equation shows that 2 moles of SO₂ produce 2 moles of SO₃, the change in concentration of SO₂ is equal to that SO₃.

Change in [SO₂] = -0.40 mol/dm³
[SO₂]eq = 0.060-0.040 = 0.020 mol/dm³

From the balanced equation, 1 mole of O₂ is consumed for every 2 moles of SO₃ formed, so:

Change in [O₂] = -½ × 0.040 = -0.020 mol/dm³
[O₂]eq = 0.050-0.020 = 0.30 mol/dm³

Answer [O₂] = 0.030 mol/dm³

Ans. Chemical equilibrium is a state in a reversible reaction where the rate of forward reaction equals the rate of backward reaction, and the concentrations of reactants and products remain constant.

Example
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

The system reaches equilibrium when ammonia is being formed and decomposed at the same rate.

Ans. A reversible reaction is one in which the products can react to form the original reactants.

Example
H₂(g) + I₂(g) ⇌ 2HI(g)

Ans. Because at equilibrium, both forward and reverse reactions continue to occur, but at equal rates, resulting in no net change in concentration of reactants and products.

Ans. No equilibrium means equal reaction rates, not equal concentrations. The amounts depend on reaction conditions and the equilibrium constant.

Ans. It states that at constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to the power of their stoichiometric coefficients.

Ans. For the reaction
aA + bB ⇌ cC + dD
The equilibrium constant (Kc) is:

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Ans. Kc is the equilibrium constant in terms of concentration. It represents the ratio of product concentrations to reactant concentrations at equilibrium.

Ans.

• Kc > 1: Products are favored (reaction goes nearly to completion).
• Kc < 1: Reactants are favored (reaction hardly proceeds).
• Kc = 1: Significant amounts of both reactants and products present.

Ans. No. Kc only depends on temperature. Changing pressure, volume, or concentrations affects equilibrium position, not Kc.

Ans.

• Homogeneous equilibrium: All reactants and products are in the same phase. Example: H₂(g) + I₂(g) ⇌ 2HI(g)
• Heterogeneous equilibrium: Reactants and products are in different phases. Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Ans. If a system at equilibrium is disturbed by changing temperature, pressure, or concentration, the system shifts its equilibrium position to oppose the change.

Ans. The system shifts to consume the added species.

Example Adding more H₂ in
H₂(g) + I₂(g) ⇌ 2HI(g)

shifts equilibrium to the right.

Ans. Increasing pressure favours the side with fewer gas molecules; decreasing favors the side with more.

Example N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Increased pressure shifts equilibrium to the right.

Ans.

• For exothermic reactions, increasing temperature shifts equilibrium to the left.
• For endothermic reactions, increasing temperature shifts equilibrium to the right.

Ans. Because the forward reaction (formation of NH₃) involves fewer gas molecules, high pressure favours ammonia production.

Ans. About 450-500°C is used to balance the rate of reaction and yield because higher temperature decreases yield (due to exothermic nature) but increases rate.

Ans. In exothermic reactions, after achieving equilibrium, if temperature is increased it favours reverse reaction. If temperature is decreased after attaining equilibrium, it favours forward reaction. In both Kc will change.

Ans. The change of catalyst will not affect or disturb the equilibrium, as catalyst helps to attain the equilibrium only by equally increasing rate of forward and reverse reaction.

Ans. In endothermic reversible reactions, increasing temperature favours forward direction. Decrease in temperature favours reverse direction. In both Kc will change.

Ans. Depends on the number of moles.
For:
aA + bB ⇌ cC + dD

Unit of Kc=(mol dm⁻³)^(c+d)-(a+b)

Ans. The reaction has very low product concentration at equilibrium; the reaction does not proceed much in the forward direction.

Ans. Because their concentrations remain constant and do not affect equilibrium.

Ans. A catalyst increases the rate of both forward and reverse reactions equally, but does not change the position of equilibrium or the value of Kc.

Ans.

Kc = [SO₃]² / [SO₂]²[O₂]

Ans. The system will shift to the right to replace the lost CO₂, favoring decomposition of CaCO₃.

Ans. Equilibrium controls the yield of products. By adjusting conditions (pressure, temperature), industries maximize product formation (e.g., ammonia, sulfuric acid).

Ans. By increasing temperature, because heat acts as a reactant and the equilibrium shifts to the right.

Ans. A mixture of reactants and products present at equilibrium state.

Ans. The relative concentrations of reactants and products at equilibrium, indicating which side is favored.

Ans. Q is calculated like Kc, but for non-equilibrium conditions. It helps predict the direction the reaction will proceed to reach equilibrium.

Ans. See Q.6 from theory.
(b) Write the expressions for Kc for the following reactions.

(i) Sn²⁺(aq) + 2Fe³⁺(aq) ⇌ Sn⁴⁺(aq) + 2Fe²⁺(aq)

Kc = [Sn⁴⁺][Fe²⁺]² / [Sn²⁺][Fe³⁺]²

(ii) Ag⁺(aq) + Fe²⁺(aq) ⇌ Fe³⁺(aq) + Ag(s)

Kc = [Fe³⁺] / [Ag⁺][Fe²⁺]

(iii) N₂(g) + O₂(g) ⇌ 2NO(g)

Kc = [NO]² / [N₂][O₂]

(iv) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)

Kc = [NO]⁴[H₂O]⁶ / [NH₃]⁴[O₂]⁵

(v) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Kc = [PCl₃][Cl₂] / [PCl₅]

Ans. (i) CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O

Kc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][CH₃CH₂OH]

(ii) 2HI ⇌ H₂ + I₂

Kc = [H₂][I₂] / [HI]²

(iii) N₂ + 3H₂ ⇌ 2NH₃

Kc = [NH₃]² / [N₂][H₂]³

Ans. The following changes occurs when:

(i) Temperature increased

• The forward dissociation (PCl₅→PCl₃+Cl₂) is endothermic.
• Increasing temperature shifts equilibrium to the right (more PCl₃, Cl₂).
• Kc increase, because increasing temperature favors the endothermic reaction for forward direction.

(ii) Volume of container decreased (pressure increased)

• Moles of gases are 1 mole at reactant side, 2 moles at product side.
• Higher pressure or lower volume shifts equilibrium to the left (towards fewer moles or product side), forming more PCl₅.
• After achieving new equilibrium, Kc remains unchanged.

(iii) Catalyst added

• It speeds up the reaction approach to equilibrium without changing the equilibrium position.
• No change in equilibrium composition, also Kc remains unchanged.

(iv) Chlorine (Cl₂) added

• Adding Cl₂ a product shifts equilibrium to the left to consume some Cl₂, forming more PCl₅.
• Kc remains unchanged, by the addition of Cl₂.

Ans. Effect of temperature change equilibrium stage

(i) If temperature is increased

• Heat is treated as a product in exothermic reactions.
• Increasing temperature shifts equilibrium to the left (reverse direction) to absorb excess heat. This decreases the yield of NH₃ so, the product formation is not favoured at high temperature.
• Equilibrium constant Kc decreases because forward reaction is less favored.

(ii) If temperature is decreased

• Lower T favors the exothermic direction (forward reaction).
• Equilibrium shifts to the right, producing more NH₃.
• Kc increases because forward reaction is more favored.

Optimum conditions (Haber's process)

• Too low a temperature gives high yield but very slow rate.
• Optimum temperature = 450 °C is used to balance rate and yield.

Ans. HI(g) ⇌ H₂(g) + I₂(g) with Kc = 0.016 at 520°C

Initial equilibrium concentrations (given)
[HI]ᵢ = 0.080 M, [H₂]ᵢ = 0.010 M, [I₂]ᵢ = 0.010 M

After adding HI [HI] becomes 0.096 M, while [H₂] = 0.010 M and [I₂] = 0.010M just after addition (before re-equilibration).

Set up as (after addition, before shift → Let x be the amount of H₂ (and I₂) formed as equilibrium shifts to the right):

2HI ⇌ H₂ + I₂

Initial (after addition) Change Equilibrium
0.096 -2x +x +x
0.010 +x 0.010 +x
0.010 +x

Write Kc expression

Kc = [H₂][I₂] / [HI]² = (0.010 + x)(0.010 + x) / (0.096 - 2x)² = 0.016

Take square root (all concentrations positive), so

0.010 + x / 0.096 - 2x = √0.016 = 0.1265

0.010 + x = 0.1264911(0.096 - 2x)
0.010 + x = 0.0121432 - 0.2529822x
x + 0.2529822x = 0.0121432 - 0.010
1.2529822x = 0.0021432
x ≈ 0.001704 M

Now plug back to get equilibrium concentrations:

• [HI]eq = 0.096 - 2x = 0.096 - 2(0.001704) = 0.0925791 M
• [H₂]eq = 0.010 + x = 0.0117104 M
• [I₂]eq = 0.0117104 M

Final answers [HI]eq = 0.09258 M, [H₂]eq = [I₂]eq = 0.01171 M

Check [H₂][I₂] / [HI]² = (0.0117104)² / (0.0925791)² = 0.016

Ans. Reaction
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Given
Kc = 4, nacetic acid,initial = 3 mol, nethanol,initial = 1 mol

No ester or water initially available,
Step 1 – Let x = moles of ester formed at equilibrium

At equilibrium
Species | Initial moles | Change | Equilibrium moles
CH₃COOH | 3 | -x | 3 - x
C₂H₅OH | 1 | -x | 1 - x
CH₃COOC₂H₅ | 0 | +x | x
H₂O | 0 | +x | x

Step 2 – Write Kc expression
Since this is a liquid-phase reaction (all in same phase), we can use molar amounts directly or concentrations if total volume is constant. The ratio remains same.

Kc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH]

Substitute equilibrium amounts 4 = x · x / (3 - x)(1 - x)

Step 3 – Solve for x
Multiply through
4(3 - x)(1 - x) = x²
4(3 - x - 3x + x²) = x²
4(3 - 4x + x²) = x²
12 - 16x + 4x² = x²
12 - 16x + 3x² = 0
3x² - 16x + 12 = 0

Step 4 – Quadratic equation
x = (16 ± √((16)² - 4(3)(12))) / (2(3))
x = (16 ± √(256 - 144)) / 6
x = (16 ± √112) / 6
x = (16 ± 10.583) / 6

Two possibilities

• x = 26.583 / 6 = 4.43 (impossible – exceeds initial ethanol moles)
• x = 5.417 / 6 = 0.903

Step 5 – Final answer x = 0.903 mol
So moles of ethyl acetate at equilibrium = 0.903 mol.

Ans. Reaction
H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)

(i) Expression for Kp

Kp = (PH₂ · PCO₂) / (PH₂O · PCO)

Numerical value of Kp (given data)
Initial: 1.00 mol H₂O and 1.00 mol CO; no H₂ or CO₂ initially
Let the equilibrium moles of H₂ be x. Then at equilibrium:

• nH₂ = x, nCO₂ = x
• nH₂O = 1 - x, nCO = 1 - x
• Total moles = (1 - x) + (1 - x) + x + x = 2 (constant)

Given 33.3% of the equilibrium mixture is hydrogen, i.e. mole fraction of H₂,

yH₂ = x/2 = 0.333, x = 0.666

So equilibrium moles
nH₂ = nCO₂ = 0.666, nH₂O = nCO = 0.334

Mole fractions
yH₂ = yCO₂ = 0.666 / 2 = 0.333,

yH₂O = yCO = 0.334 / 2 = 0.167

Partial pressures Pi = yi · Ptotal. Substitute into Kp:

Kp = (yH₂ · P)(yCO₂ · P) / (yH₂O · P)(yCO · P)
= (yH₂ · P)(yCO₂ · P) / (yH₂O · P)(yCO · P)
= (yH₂ · yCO₂) / (yH₂O · yCO)

(the total pressure P cancels because there are two factors of P in numerator and denominator)

Now plug values

Kp = (0.333)(0.333) / (0.167)(0.167) = 0.110889 / 0.027889 ≈ 3.98(≈ 4.0)

A quicker algebraic route (same result): since yH₂ = x/2 and yH₂O = (1-x)/2,

Kp = (x/2)² / ((1-x)/2)² = (x/(1-x))²

With x = 0.666 gives Kp = (0.666/0.334)² = 3.98

Unit of Kp
When Δn = 0, Kp is dimensionless (unitless). Kp has no unit.