Grade 12 Physical Sciences · Chemical Change

Calculate reaction rate

Calculate reaction rate as change in amount over change in time (20 cm³ of gas made in 10 s is 2 cm³/s).

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Question 5.4 · May/June 2025 · Paper 2 · National (DBE) · 6 marks

The reaction between magnesium carbonate pellets,

MgCO_{3}

(s), and EXCESS dilute hydrochloric acid, HCℓ(aq), is used to investigate the effect of concentration on the rate of a reaction.

MgCO_{3} (s) + 2 HC ℓ (aq) \to MgC ℓ_{2} (aq) + H_{2} O (ℓ) + CO_{2} (g)

The apparatus used for this investigation is shown below. Diagram for this question

5.4 For curve B, calculate the average rate at which

CO_{2}

(g) is produced for the first

120

s in dm

^{3}

·s

^{- 1}

. The molar gas volume is

24, 5

^{3}

·mol

^{- 1}

.(6)

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More past paper questions: calculate reaction rate

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3. Question 5.3.2 · May/June 2023 · Paper 2 · National (DBE) · 4 marks
Consider the following decomposition reaction that takes place in a sealed $2$ dm $^{3}$ container:

$2 N_{2} O_{5} (g) \to 4 NO_{2} (g) + O_{2} (g)$

The graph below shows how the concentrations of $N_{2} O_{5}$ (g) and $NO_{2}$ (g) change with time.
5.3 Calculate the:
5.3.2 Average rate of production of $O_{2}$ (g) in mol·dm $^{- 3}$ ·s $^{- 1}$ in $700$ s(4)
Show the memo (4 ticks)Memo
1. ✓
  Substitute the change in concentration into the rate formulaAve rate $= - \frac{Δ c ( N _{2} O _{5} )}{Δ t} = - \frac{( 60 \times 1 0 ^{- 4} - 200 \times 1 0 ^{- 4} )}{...}$
  1 mark · Calculate reaction rate
2. ✓
  Substitute time into the rate formula $= - \frac{( 60 \times 1 0 ^{- 4} - 200 \times 1 0 ^{- 4} )}{700 - 0} = 2 \times 1 0^{- 5}$ mol·dm $^{- 3}$ ·s $^{- 1}$
  1 mark · Calculate reaction rate
3. ✓
  Use mole ratiorate( $O_{2}$ ) $= \frac{1}{2}$ rate( $N_{2} O_{5}$ ) $= \frac{1}{2} (2 \times 1 0^{- 5})$
  1 mark · Use the mole ratio
4. ✓
  Final answer $= 1 \times 1 0^{- 5}$ mol·dm $^{- 3}$ ·s $^{- 1}$
  1 mark · Use the mole ratio
4. Question 5.2.2 · November 2025 · Paper 2 · National (DBE) · 5 marks
5.2 Carbon monoxide, CO(g), reacts with oxygen, $O_{2}$ (g), to form carbon dioxide, $CO_{2}$ (g), in a sealed container according to the balanced equation:

$2 CO(g) + O_{2} (g) \to 2 CO_{2} (g)$

The graph below shows the concentration of $O_{2}$ (g) versus time.

Answer the questions that follow.
5.2.2 The reaction takes place in a $3$ dm $^{3}$ container. Calculate the average rate (in mol·s $^{- 1}$ ) at which $CO_{2}$ (g) is formed in the first 10 s.(5)
Show the memo (5 ticks)Memo
1. ✓
  (a) Calculate the change in concentrationRate $= - \frac{Δ c ( O _{2} )}{Δ t}$
  $= - \frac{( 0 , 15 - 0 , 27 )}{\dots}$
  1 mark · Calculate reaction rate
2. ✓
  (b) Substitute 10 s $= - \frac{( 0 , 15 - 0 , 27 )}{10 - 0}$
  $= 0, 012$ mol·dm $^{- 3}$ ·s $^{- 1}$
  1 mark · Calculate reaction rate
3. ✓
  (c) Multiply by 3 dm3Rate ( $O_{2}$ ) in mol·s $^{- 1}$ $= c V = (0, 012) (3) = 0, 036$ mol·s $^{- 1}$
  1 mark · Calculate concentration of a solution
4. ✓
  (d) Use the mol ratioRate ( $CO_{2}$ ) $= 2 \times$ rate ( $O_{2}$ ) $= 2 \times 0, 036$
  1 mark · Use the mole ratio
5. ✓
  (e) Final answer $= 0, 072$ (mol·s $^{- 1}$ )
  1 mark · Use the mole ratio
5. Question 5.4.3 · November 2022 · Paper 2 · National (DBE) · 5 marks
Three experiments, A, B and C, are carried out to investigate some of the factors that affect the rate of decomposition of hydrogen peroxide, $H_{2} O_{2} (ℓ)$ .

The balanced equation for the reaction is:

$2 H_{2} O_{2} (ℓ) \to 2 H_{2} O (ℓ) + O_{2} (g)$

Identical samples of hydrogen peroxide are used in each experiment.

The conditions used in each experiment are summarised in the table below.
5.4 The graph below, NOT drawn to scale, is obtained for the mass of oxygen gas produced over a period of time in experiment A.

Use the information in the graph to answer the following questions:
The average rate of decomposition of hydrogen peroxide is $2, 1 \times 1 0^{- 3}$ mol·s $^{- 1}$ .
5.4.3 Calculate the value of time t on the graph.(5)
Show the memo (5 ticks)Memo
1. ✓
  Substitute 0,9 g $n (O_{2}) = \frac{m}{M} = \frac{0 , 9}{M}$
  1 mark · Calculate number of moles
2. ✓
  Substitute 32 g·mol⁻¹ $n (O_{2}) = \frac{0 , 9}{32} = 0, 028$ mol $(0, 0281)$
  1 mark · Calculate number of moles
3. ✓
  Mole ratio $n (H_{2} O_{2}) = 2 n (O_{2}) = 2 (0, 028) = 0, 056$ mol
  1 mark · Use the mole ratio
4. ✓
  Substitute into the rate formula $rate = \frac{Δ n}{Δ t}$
  
  $2, 1 \times 1 0^{- 3} = \frac{0 , 056 - 0}{Δ t}$
  1 mark · Calculate reaction rate
5. ✓
  Final answer $Δ t = 26, 67$ (s), so $t = 26, 67$ s
  1 mark · Calculate reaction rate
6. Question 5.4.1 · May/June 2022 · Paper 2 · National (DBE) · 6 marks
Learners use the reaction of $MgCO_{3}$ (s) with EXCESS dilute HC $ℓ$ (aq) to investigate the relationship between temperature and the rate of a chemical reaction.

The balanced equation for the reaction is:

$MgCO_{3} (s) + 2 HC ℓ (aq) \to MgC ℓ_{2} (aq) + CO_{2} (g) + H_{2} O (ℓ)$

The results obtained are represented in the graph below.
5.4 The learners obtained the graph above using $5$ g $MgCO_{3}$ (s) with EXCESS HC $ℓ$ at $40$ °C.

Calculate the:
5.4.1 Time taken for the reaction to run to completion(6)
Show the memo (6 ticks)Memo
1. ✓
  Formula n = m/M $n (MgCO_{3}) = \frac{m}{M}$
  1 mark · Calculate number of moles
2. ✓
  Substitution of 84 g·mol⁻¹ $= \frac{5}{84} = 0, 06$ mol ( $0, 0595$ mol)
  1 mark · Calculate number of moles
3. ✓
  Mole ratio $n (CO_{2})_{produced} = n (MgCO_{3}) = 0, 06$ mol
  1 mark · Use the mole ratio
4. ✓
  Substitution of 44 g·mol⁻¹ $n (CO_{2}) = \frac{m}{M}$
  $0, 06 = \frac{m}{44}$
  $m (CO_{2}) = 2, 64$ g
  1 mark · Calculate number of moles
5. ✓
  Substitution of 0,5 in the rate equationAve rate $= \frac{Δ m ( CO _{2} )}{Δ t}$
  $0, 5 = \frac{2 , 64}{Δ t}$
  1 mark · Calculate reaction rate
6. ✓
  Final answer $Δ t = 5, 28$ min
  1 mark · Calculate reaction rate

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