DP IB Chemistry: HL

Topic Questions

IBChemistry: HLDPTopic Questions17. Equilibrium (HL only)17.1 The Equilibrium Law

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First teaching 2014

Last exams 2024

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17.1 The Equilibrium Law

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1
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Which is the correct Kc expression for the reaction between hydrogen and fluorine?


H2 (g) + F2 (g) ⇌ 2HF (g)

  • fraction numerator left square bracket HF right square bracket squared over denominator left square bracket straight H subscript 2 right square bracket space left square bracket straight F subscript 2 right square bracket end fraction

  • fraction numerator left square bracket HF right square bracket over denominator left square bracket straight H subscript 2 right square bracket space left square bracket straight F subscript 2 right square bracket end fraction

  • fraction numerator 2 space left square bracket HF right square bracket over denominator left square bracket straight H subscript 2 right square bracket space left square bracket straight F subscript 2 right square bracket end fraction

  • fraction numerator left square bracket straight H subscript 2 right square bracket space left square bracket straight F subscript 2 right square bracket over denominator left square bracket HF right square bracket squared end fraction

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2
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The Kc expression for the following reaction between hydrogen and iodine is shown 

H2 (g) + I2 (g) ⇌ 2HI (g)

K subscript straight C equals fraction numerator left square bracket HI right square bracket squared over denominator left square bracket straight H subscript 2 right square bracket space left square bracket straight I subscript 2 right square bracket end fraction

At equilibrium there were 0.234 moles of HI, 0.150 moles of H2 and 0.025 moles of I2.


Which is the correct Kc expression for the reaction between hydrogen and fluorine?

  • fraction numerator left square bracket 0.234 right square bracket squared over denominator left square bracket 0.15 right square bracket space left square bracket 0.025 right square bracket end fraction

  • fraction numerator left square bracket 0.234 right square bracket over denominator left square bracket 0.15 right square bracket space left square bracket 0.025 right square bracket end fraction

  • fraction numerator left square bracket 0.15 right square bracket space left square bracket 0.025 right square bracket over denominator left square bracket 0.234 right square bracket squared end fraction

  • fraction numerator left square bracket 0.15 right square bracket space left square bracket 0.025 right square bracket over denominator left square bracket 0.234 right square bracket end fraction

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3
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Nitrosyl chloride decomposes into nitrogen monoxide and chlorine according to the following equation. The forward reaction is endothermic 

2NOCl (g) ⇌ 2NO (g) + Cl2 (g)


Which change in condition would change the value for Kc?

  • Decreasing the pressure

  • Adding a catalyst

  • Increasing the temperature

  • Increasing the pressure

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4
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Which of the following rows correctly describes Kc and ΔG for a reaction where the products are favoured?

  Kc ΔG
A. > 1 < 1
B. > 1 < 0
C. > 0 > 0
D. > 0 > 1

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5
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At 300 K, iron oxidises according to the following equation:

2Fe (s) + 3 over 2O2 (g) ⇌ Fe2O3 (s)

The standard Gibbs free energy change for this reaction is -743.05 kJ mol-1.

The quantitative relationship between the standard Gibbs free energy change, temperature and the equilibrium constant is:

ΔG = –RT ln K

Which expression is a correct step towards calculating the value of the equilibrium constant? (R = 8.31 J K–1 mol–1)

  • ln space K equals fraction numerator 8.31 cross times 573 over denominator negative 743.05 cross times 10 cubed end fraction

  • ln space K equals fraction numerator 8.31 cross times 300 over denominator negative 743.05 cross times 10 cubed end fraction

  • ln space K equals fraction numerator negative 743.05 over denominator 8.31 cross times 300 end fraction

  • ln space K equals fraction numerator negative 743.05 cross times 10 cubed over denominator 8.31 cross times 300 end fraction

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1
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The Haber process is a key step in the manufacture of fertilisers:

N2 (g) + 3H2 (g) rightwards harpoon over leftwards harpoon 2NH3 (g)         ΔH = -ve

Which is correct about the effect of increasing temperature for this reaction?

  Effect on equilibrium position Effect on Kc
A. Shifts left No change
B. Shifts right No change
C. Shifts right Increase
D. Shifts left Decrease

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2
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Which equation represents a reaction where the number of moles alone can not be used to calculate the value of Kc?

  • CH3CH2OH (aq) + CH3COOH (aq) rightwards harpoon over leftwards harpoon CH3CH2OCOCH3 (aq)  + H2O (l)

  • H2 (g) + I2 (g) rightwards harpoon over leftwards harpoon 2HI (g)

  • 2SO2 (g) + O2 (g) rightwards harpoon over leftwards harpoon 2SO3 (g)

  • N2 (g) + O2 (g) rightwards harpoon over leftwards harpoon 2NO (g)

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3
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When 0.20 mol NO, 0.08 mol H2 and 0.10 mol of H2O are placed in a 1.0 dm3 flask, the following equilibrium is established:

2NO (g) + 2H2 (g) rightwards harpoon over leftwards harpoon N2 (g) + 2H2O (g)

At equilibrium, the concentration of H2 (g) was found to be 0.02 mol dm-3.

What is the correct calculation to work out Kc?

  • Kc = fraction numerator stretchy left square bracket 0.14 stretchy right square bracket squared stretchy left square bracket 0.02 stretchy right square bracket squared over denominator stretchy left square bracket 0.03 stretchy right square bracket to the power of blank stretchy left square bracket 0.16 stretchy right square bracket squared end fraction

  • Kc = fraction numerator open square brackets 0.03 close square brackets open square brackets 0.16 close square brackets squared over denominator open square brackets 0.14 close square brackets squared open square brackets 0.02 close square brackets squared end fraction

  • Kc = fraction numerator begin mathsize 14px style stretchy left square bracket 0.03 stretchy right square bracket stretchy left square bracket 0.06 stretchy right square bracket squared end style over denominator begin mathsize 14px style stretchy left square bracket 0.06 stretchy right square bracket squared stretchy left square bracket 0.06 stretchy right square bracket squared end style end fraction

  • Kc = fraction numerator begin mathsize 14px style stretchy left square bracket 0.06 stretchy right square bracket stretchy left square bracket 0.16 stretchy right square bracket squared end style over denominator begin mathsize 14px style stretchy left square bracket 0.14 stretchy right square bracket squared stretchy left square bracket 0.04 stretchy right square bracket squared end style end fraction

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4
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Nitrogen dioxide can form a dimer that can also break back down again as part of a reversible reaction:

N2O4 (g) rightwards harpoon over leftwards harpoon NO2 (g)         ΔH = +ve

The reaction reaches an equilibrium at temperature T, where Kc = 1

What is true for a higher temperature, T2?

  Kc value ΔGθ value
A. Increases Increases
B. Decreases Increases
C. Decreases Decreases
D. Increases Decreases

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5
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Which would be the correct way to plot a graph and then calculate ΔGθ from experimental data of Kc and temperature values?

ΔGθ = -RT ln K

  y-axis x-axis ΔGθ =
A. 1 / T ln K -R x gradient
B. ln K 1 / T -R x gradient
C. ln K 1 / T R / gradient
D. 1 / T ln K R / gradient

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11 mark

When gaseous dinitrogen pentoxide, N2O5 (g), decomposes at 358 K, the following equilibrium is established:

2N2O5 (g)  ⇌ 4NO2 (g) + O2 (g)

2.0 mol of N2O5 (g) were placed in a 1.0 dm3 container and allowed to reach equilibrium. At equilibrium 1.0 mol of N2O5 (g) were present. What is the value of Kc?

  • 0.125

  • 1

  • 2

  • 8

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21 mark

Consider the following reversible reaction:

3O2 (g) ⇌ 2O3 (g) 

What is the value of Kc when the equilibrium concentrations are [O2] = 4.0 mol dm-3 and [O3] = 4.0 mol dm-3 ?

  • 0.25

  • 4

  • 16

  • 64

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31 mark

Which of the following will shift the position of equilibrium to the right in the reaction shown? 

2N2O5 (g)  ⇌ 4NO2 (g) + O2 (g)     ΔH = +219.2 kJ 

  1.  Decreasing the concentration of NO2 (g)
  2.  Decreasing the temperature
  3.  Decreasing the pressure

  • I and II only

  • I and III only

  • II and III only 

  • I, II and III

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41 mark

Hydrogen iodide decomposes to form hydrogen and iodine vapour.

2HI (g)  ⇌ H2 (g) + I2 (g)

What is the effect of decreasing the volume of the equilibrium mixture at constant temperature?

  • The amount of H2 (g) remains the same but its concentration decreases

  • The forward reaction is favoured

  • The backward reaction is favoured

  • The value of Kc  remains unchanged

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51 mark

A mixture of 0.40 mol of SO2 (g) and 0.40 mol of O2 (g) was placed in a 1 dm3 container. The following equilibrium took place:

2SO2 (g)  + O2 (g) ⇌  2SO3 (g) 

At equilibrium the mixture contained 0.25 mol of O2 (g) .How many moles of SO2 (g) and SO3 (g) were present at equilibrium?

 

SO2 (g) / mol

SO3 (g) / mol

A

0.25

0.15

B

0.30

0.15

C

0.10

0.30

D

0.25

0.30

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