Calorimetry Lab: Measuring Enthalpy of Reaction量热实验：测定反应焓变

AP Science Practice 5 · Mathematical Routines~35 min

Learning goals

  • Find q from the solution’s mass, specific heat and temperature change, then ΔH per mole of limiting reactant, with the correct sign
  • Find ΔT by extrapolating a temperature–time curve, and measure and use a calorimeter’s heat capacity when needed
  • State the usual assumptions and sources of error in calorimetry

Before you start

  • Heat capacity and calorimetry (U6.4)
  • Introduction to enthalpy of reaction (U6.6)
  • Experimental design and error analysis (4.3.5)

The coffee-cup calorimeter

The reaction runs in a covered foam cup (at constant pressure) while a thermometer records the solution’s temperature. The calculation has three steps:

qsoln=mcΔTqrxn=−qsolnΔH=qrxnnlimiting\begin{gathered} q_{\text{soln}} = mc\Delta T \qquad q_{\text{rxn}} = -q_{\text{soln}} \\ \Delta H = \frac{q_{\text{rxn}}}{n_{\text{limiting}}} \end{gathered}
  • m is the mass of the whole solution (count both solutions; when dissolving a solid, include the solid).
  • If the temperature rises, the solution absorbs heat and the reaction releases it, so ΔH < 0; if the temperature falls, ΔH > 0.
  • If the calorimeter itself absorbs a significant amount of heat, add CcalΔT to qsoln.

Usual assumptions

  • A dilute solution has a density of 1.00 g/mL and the specific heat of water, 4.18 J/(g·°C).
  • No heat is lost to the surroundings, and the calorimeter itself takes up too little heat to matter (or that heat is measured separately).

Temperature–time curves and extrapolation

Heat release (or absorption) takes a little time, and meanwhile the solution exchanges heat with its surroundings, so the highest (or lowest) thermometer reading falls short of the true value. The fix: record the temperature over time before and after mixing, then extrapolate the later, nearly straight section back to the moment of mixing, and use that temperature to find ΔT.

Calorimetry lab: finding ΔT and ΔH from a temperature–time curve

060120240360Time (s)Temperature (°C)

Example 1: ΔH of neutralization

50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH are mixed, and extrapolation gives ΔT = 6.60 °C. Take the solution’s density as 1.00 g/mL and its specific heat as 4.18 J/(g·°C), and ignore heat absorbed by the calorimeter. Find ΔH of neutralization.

Step-by-step solution

  1. Heat absorbed by the solution. The solution totals 100.0 mL, or 100.0 g:

    qsoln=(100.0)(4.18)(6.60)=2.76×103 J\begin{aligned} q_{\text{soln}} &= (100.0)(4.18)(6.60) \\ &= 2.76 \times 10^{3}\ \text{J} \end{aligned}

  2. Heat released by the reaction. qrxn = −2.76 × 10³ J = −2.76 kJ.

  3. Moles of water formed. HCl and NaOH are 0.0500 mol each and react exactly, forming 0.0500 mol of H₂O.

  4. ΔH.

    ΔH=−2.76 kJ0.0500 mol=−55.2 kJ/mol\Delta H = \frac{-2.76\ \text{kJ}}{0.0500\ \text{mol}} = \boxed{-55.2\ \text{kJ/mol}}

    That’s within about 1% of the −55.8 kJ/mol calculated from enthalpies of formation.

Example 2: measuring a calorimeter’s heat capacity

In a calorimeter, 50.0 g of water at 55.0 °C is mixed with 50.0 g of water at 22.0 °C, reaching a final temperature of 37.6 °C. Find the heat capacity of the calorimeter, Ccal.

Step-by-step solution

  1. Heat released by the hot water.

    qlost=(50.0)(4.18)(55.0−37.6)=3637 J\begin{aligned} q_{\text{lost}} &= (50.0)(4.18)(55.0 - 37.6) \\ &= 3637\ \text{J} \end{aligned}

  2. Heat absorbed by the cold water.

    qgained=(50.0)(4.18)(37.6−22.0)=3260 J\begin{aligned} q_{\text{gained}} &= (50.0)(4.18)(37.6 - 22.0) \\ &= 3260\ \text{J} \end{aligned}

  3. The difference went into the calorimeter, which warmed from 22.0 °C to 37.6 °C along with the cold water:

    Ccal=3637−326037.6−22.0=24 J/∘C\begin{aligned} C_{\text{cal}} &= \frac{3637 - 3260}{37.6 - 22.0} \\ &= \boxed{24\ \text{J}/{}^\circ\text{C}} \end{aligned}
Practice: from what you observe, what is the sign of ΔH?
  1. Hydrochloric acid and NaOH solution are mixed and the temperature rises
  2. NH₄NO₃ dissolves in water and the solution gets cold
  3. CaCl₂ dissolves in water and the solution gets warm
  4. KNO₃ dissolves in water and the temperature drops
  5. Magnesium ribbon is put in hydrochloric acid and the solution warms
  6. Baking soda and vinegar are mixed and the temperature drops
  7. Ice is added to water in a calorimeter and the water cools
  8. Solid NaOH dissolves in water and the solution gets noticeably hot

Common mistakes

Mistake 1: “qsoln is ΔH.” qsoln is the heat the solution absorbs; the reaction’s q has the opposite sign, and ΔH is that divided by the moles of limiting reactant, in kJ/mol.

Mistake 2: “The solution’s mass is just one of the solutions.” m is the mass of the whole mixed solution. Mixing 50.0 mL of acid with 50.0 mL of base gives m = 100.0 g; when dissolving a solid, include the solid’s mass too.

Mistake 3: “The highest thermometer reading gives the most accurate ΔT.” Heat leaks away while the reaction happens, so the highest reading is too low. Extrapolating the temperature–time curve back to the mixing time gives a more accurate ΔT.

Quick check

Why is a covered foam cup used for calorimetry instead of a glass beaker?

A reaction causes the solution to absorb 2.76 kJ of heat. What is q for the reaction?

50.0 mL of 1.00 M HCl is mixed with 25.0 mL of 1.00 M NaOH. What value of n should be used to calculate ΔH?

Why is the temperature–time curve extrapolated back to the mixing time?

100.0 g of solution warms by 5.0 °C (specific heat 4.18 J/(g·°C)). How much heat does it absorb?

Why is the density of a dilute solution usually taken as 1.00 g/mL?

AP-style practice

AP-style free response

A student drops 0.1215 g of magnesium ribbon into 100.0 mL of 1.00 M hydrochloric acid in a coffee-cup calorimeter: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g). The highest thermometer reading is 5.32 °C above the starting temperature. Take the solution’s density as 1.00 g/mL and its specific heat as 4.18 J/(g·°C) (Mg 24.31 g/mol).

(a) Show by calculation which reactant is limiting.

(b) Calculate the heat absorbed by the solution.

(c) Calculate ΔH for the reaction, in kJ per mole of Mg.

(d) ΔH° calculated from enthalpies of formation is −466.9 kJ/mol. Calculate the percent error of your result in (c), and suggest one reason |ΔH| is too small.

(e) Explain how temperature–time data could give a more accurate ΔT.

(f) The magnesium ribbon had a coating of MgO that the student didn’t remove before weighing. Is the calculated |ΔH| too large or too small? Explain.

Model answer and scoring

(a) n(Mg) = 0.1215 ÷ 24.31 = 5.00 × 10⁻³ mol; n(H⁺) = 0.1000 L × 1.00 M = 0.100 mol, and only 1.00 × 10⁻² mol of H⁺ is needed, so Mg is limiting.

(b) qsoln = 100.0 g × 4.18 J/(g·°C) × 5.32 °C = 2.22 × 10³ J.

(c) ΔH = −2.22 kJ ÷ (5.00 × 10⁻³ mol) = −445 kJ/mol.

(d) Percent error = |−445 − (−466.9)| ÷ 466.9 × 100% = 4.7%. Possible reasons: heat lost to the air or absorbed by the calorimeter; using the highest reading without extrapolating.

(e) Record the temperature over time before and after adding the magnesium, plot temperature against time, and extrapolate the nearly straight section after the peak back to the time the magnesium was added; ΔT is that temperature minus the starting temperature.

(f) Too small. Part of the weighed mass is MgO, so there is less Mg than assumed and less heat is released; meanwhile n(Mg), calculated from the full mass, is too large. With q too small and n too large, |ΔH| = q/n is too small.

Scoring: (a) moles of both reactants and the comparison; (b) the value of q; (c) ΔH with its negative sign; (d) the percent error and one reasonable cause; (e) extrapolating to the mixing time; (f) too small, with the chain of reasoning.

Summary

  • qsoln = mcΔT (m is the whole solution’s mass), qrxn = −qsoln, and ΔH = qrxn ÷ n(limiting reactant).
  • A temperature rise means ΔH < 0 and a drop means ΔH > 0; extrapolating the temperature–time curve corrects for heat loss.
  • Usual assumptions: density 1.00 g/mL, specific heat 4.18 J/(g·°C), no heat loss; heat loss usually makes |ΔH| too small.

AP tip: In calorimetry, points are most often lost on the sign and the units. A complete final line reads “ΔH = −445 kJ/mol,” with the negative sign and kJ/mol.

Sources: standard enthalpy changes for neutralization, dissolving NH₄NO₃, and Mg with hydrochloric acid calculated from enthalpies of formation in OpenStax Chemistry 2e Appendix G; the experimental data are illustrative.

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