Mr Cole Chemistry

Kc Questions

Chemical Equilibria

Worksheet

Kc Calculations

2.4 moles of phosphorous trichloride and 3.2 moles of chlorine gas were mixed in a container that had a volume of 200cm³. They were left to reach equilibrium at a constant temperature. Once at equilibrium, there were 1.1 moles of phosphorous pentachloride.
a) Write the symbol equation for the reaction.
PCl3(g) + Cl2(g) ⇌ PCl5(g)
b) Write the expression for the Kc value and determine the units.

Expression:

$$K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]}$$

Units:

$$\frac{(mol~dm^{-3})}{(mol~dm^{-3})(mol~dm^{-3})} = \frac{1}{(mol~dm^{-3})} = mol^{-1}dm^{3}$$

c) Calculate the value of Kc for the reaction.

First, convert volume to dm3: 200 cm3 = 0.20 dm3.

Species PCl3 Cl2 PCl5
Initial Moles 2.4 3.2 0.0
Change -1.1 -1.1 +1.1
Eq. Moles 1.3 2.1 1.1
Conc (c=n/V) 6.5 10.5 5.5

$$K_c = \frac{5.5}{(6.5 \times 10.5)} = \frac{5.5}{68.25} = 0.08058…$$

Answer: 0.081 mol-1dm3 (2 s.f.)

d) Given that ΔH = -420kJmol⁻¹, state and explain the effect of increasing the temperature on Kc.

The forward reaction is exothermic (negative ΔH). Increasing the temperature favors the endothermic (reverse) reaction to absorb the added heat. This shifts the equilibrium to the left, decreasing the concentration of products and increasing reactants. Therefore, the value of Kc decreases.

e) State and explain the effect of decreasing the volume of the container on the value of Kc.

Decreasing the volume would increase the pressure, shifting equilibrium to the side with fewer gas moles (the right). However, Kc remains constant because it is temperature-dependent only.

12.0 moles of hydrogen and 12.0 moles of bromine were mixed in a container that was 500cm³. Once they reached equilibrium, there were 2.3 moles of hydrogen left.
a) Write the symbol equation for the reaction.
H2(g) + Br2(g) ⇌ 2HBr(g)
b) Write the expression for the Kc value and determine the units.

Expression:

$$K_c = \frac{[HBr]^2}{[H_2][Br_2]}$$

Units:

$$\frac{(mol~dm^{-3})^2}{(mol~dm^{-3})(mol~dm^{-3})} = \text{No Units}$$

c) Calculate the value of Kc for the reaction.

Since the moles of gas are equal on both sides, volume cancels out. We can calculate using moles directly, but we will use the ICE table method.

Species H2 Br2 2HBr
Initial Moles 12.0 12.0 0.0
Change -9.7 -9.7 +19.4
Eq. Moles 2.3 2.3 19.4

$$K_c = \frac{(19.4)^2}{(2.3 \times 2.3)} = \frac{376.36}{5.29} = 71.14…$$

Answer: 71 (2 s.f.)

d) State and explain what would happen to the position of equilibrium if you were to increase the pressure.

Increasing pressure has no effect on the position of equilibrium. This is because there are 2 moles of gas on the reactant side and 2 moles of gas on the product side. Pressure only affects equilibrium if there is a difference in gas moles.

4.00 moles of sulfur dioxide and 4.00 moles of oxygen were mixed in a 2.00 dm³ container. The concentration of sulfur trioxide gradually increased until it remained constant at 1.35 moldm⁻³.
a) Write the symbol equation for the reaction.
2SO2(g) + O2(g) ⇌ 2SO3(g)
b) Calculate the value of Kc for the reaction. Include the correct unit in your answer.

Initial Concentrations: [SO2] = 4.00/2.00 = 2.00 M; [O2] = 2.00 M.

Species (conc) 2SO2 O2 2SO3
Initial 2.00 2.00 0.00
Change -1.35 -0.675 +1.35
Equilibrium 0.65 1.325 1.35

$$K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} = \frac{(1.35)^2}{(0.65)^2(1.325)} = \frac{1.8225}{0.5598…} = 3.255…$$

Answer: 3.26 mol-1dm3 (3 s.f.)

c) A vanadium oxide catalyst was added to the mixture. State and explain the impact.

Rate: The rate at which equilibrium is attained increases because the catalyst provides an alternative pathway with lower activation energy, increasing the frequency of successful collisions.

Position: The position of equilibrium remains unchanged because the catalyst increases the rate of the forward and reverse reactions equally.

250cm³ of 0.5moldm⁻³ iron (III) chloride and 250cm³ of 0.5moldm⁻³ ammonium thiocyanate react to make iron (III) thiocyanate and ammonium chloride in a beaker. At equilibrium, the concentration of ammonium chloride is 0.12 moldm⁻³.
a) Write the symbol equation for the reaction. Hint: this is not a redox reaction.
FeCl3 + 3NH4SCN ⇌ Fe(SCN)3 + 3NH4Cl
b) Calculate the value of Kc for the reaction. Include the correct unit in your answer.

Total Volume = 500 cm3 = 0.5 dm3.

Initial Moles: n(FeCl3) = 0.5 × 0.25 = 0.125 mol. n(NH4SCN) = 0.5 × 0.25 = 0.125 mol.

Eq Moles of NH4Cl: n = c × v = 0.12 × 0.5 = 0.06 mol.

Species FeCl3 3NH4SCN Fe(SCN)3 3NH4Cl
Ratio 1 3 1 3
Initial Mol 0.125 0.125 0 0
Change -0.02 -0.06 +0.02 +0.06
Eq. Mol 0.105 0.065 0.02 0.06
Eq. Conc 0.21 0.13 0.04 0.12

$$K_c = \frac{[Fe(SCN)_3][NH_4Cl]^3}{[FeCl_3][NH_4SCN]^3}$$

$$K_c = \frac{(0.04)(0.12)^3}{(0.21)(0.13)^3} = \frac{0.00006912}{0.000461…} = 0.149…$$

Answer: 0.15 (No Units)

100g of bismuth (III) chloride is dissolved in 2.40 dm³ of water. At equilibrium, the solution was found to contain HCl at a concentration of 0.20 moldm⁻³.
BiCl₃ (aq) + H₂O (l) ⇌ BiOCl (s) + 2HCl (aq)
a) What is the value for Kc for this reaction.

Important: H2O(l) and BiOCl(s) are not included in the Kc expression.

$$K_c = \frac{[HCl]^2}{[BiCl_3]}$$

Step 1: Initial Moles BiCl3
Mr(BiCl3) = 209 + (3 × 35.5) = 315.5 g/mol.
Moles = 100 / 315.5 = 0.317 mol.

Step 2: ICE Table (Work in Moles)
Eq Moles HCl = 0.20 mol dm-3 × 2.40 dm3 = 0.48 mol.

Species BiCl3 2HCl
Initial Mol 0.317 0
Change (1:2) -0.24 +0.48
Eq. Mol 0.077 0.48
Conc (n/2.4) 0.0321 0.20

$$K_c = \frac{(0.20)^2}{0.0321} = 1.246…$$

Answer: 1.2 mol dm-3

b) What impact would adding NaOH to the solution have on the position of equilibrium and on the Kc.

Equilibrium Position: NaOH reacts with HCl (neutralization), lowering the concentration of HCl. The equilibrium shifts to the right to replace the lost product.

Kc: No impact. Kc is constant at a constant temperature.

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