Physical Chemistry
8 marks · NormalA student investigates the thermal decomposition of calcium carbonate and the effect of acid concentration on the rate of this process.
Thermal decomposition of calcium carbonate is carried out: During an experiment, 2.50 g of calcium carbonate is heated, causing a temperature decrease in the surroundings from 25.0 °C to 18.1 °C. Assume all absorbed energy came from a nearby 150 g beaker of water (specific heat capacity ). Calculate the heat absorbed (Q) by the reaction mixture.
Calculate the molar enthalpy change (ΔH) for the decomposition of calcium carbonate, in kJ/mol. (M_r of )
In a separate rate experiment, the same mass of calcium carbonate is added to 50 cm³ of hydrochloric acid at two different concentrations (1.0 mol/dm³ and 0.5 mol/dm³). Predict and explain the effect of decreasing the concentration of the acid on the rate of CO₂ production.
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(a) Q = 150 × 4.18 × (18.1 − 25.0) = 150 × 4.18 × (−6.9) = −4320 J (heat absorbed). J · must show Q = mcΔT, ΔT negative
(b) n = 2.50 / 100 = 0.025 mol; ΔH = (−4.32 kJ) / 0.025 mol = −173 kJ/mol (to 3 sf, positive value since decomposition is endothermic, ΔH = +173 kJ/mol). kJ/mol · must show calculated moles, ΔH sign correct
(c) Lowering the acid concentration decreases the rate of CO₂ production. This is because there are fewer acid particles per cm³, so collisions with CaCO₃ particles happen less frequently, leading to a slower reaction (rate). · must show states rate decreases, explains fewer collisions per second, links to concentration
Also accept: Q calculation using sign conventions
Common mistakes: using wrong value for ΔT; incorrect units
Show mark scheme and worked solution
Marking points
A1 (a) · 1 mark · Uses Q = mcΔT with correct values.
A2 (a) · 1 mark · Correct substitution and negative ΔT; Q = −4320 J (−4.32 kJ).
B1 (b) · 1 mark · Calculates moles of CaCO₃ correctly. (follow-through)
B2 (b) · 1 mark · Converts Q to kJ and divides by moles for ΔH. (follow-through)
B3 (b) · 1 mark · Correct sign and units for molar ΔH. (follow-through)
C1 (c) · 1 mark · Predicts lower rate with lower concentration.
C2 (c) · 1 mark · Explains in terms of fewer acid particles per cm³/fewer collisions.
C3 (c) · 1 mark · Links collision theory or collision frequency to rate.
Worked solution
a) Q = mcΔT = 150 × 4.18 × (18.1 – 25.0) = 150 × 4.18 × (−6.9) = 150 × (−28.84) = −4326 J ≈ −4320 J. (Heat absorbed by reaction, negative as water cooled.)
b) Moles CaCO₃ = 2.50 / 100 = 0.025 mol. Q = −4.32 kJ for 0.025 mol. ΔH for 1 mol = (−4.32 kJ) / 0.025 mol = −173 kJ/mol. However, the reaction is endothermic (heat absorbed), so ΔH = +173 kJ/mol.
c) Lower acid concentration means fewer acid particles per volume. This reduces frequency of collisions with solid CaCO₃ particles so rate of CO₂ production is slower (rate decreases with concentration, as explained by collision theory).