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Reactivity 1. What drives chemical reactions? - Measuring enthalpy change

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Enthalpy change (DeltaH\\Delta H) is the heat energy exchanged between a system and its surroundings at constant pressure.

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The system is the chemical reaction itself, while the surroundings include the solvent, container, and the air.

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Exothermic reactions release heat to the surroundings; the temperature of the surroundings increases, and DeltaH\\Delta H is negative (−DeltaH-\\Delta H).

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Endothermic reactions absorb heat from the surroundings; the temperature of the surroundings decreases, and DeltaH\\Delta H is positive (+DeltaH+\\Delta H).

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Specific heat capacity (cc) is the energy required to raise the temperature of 1textg1\\text{ g} of a substance by 1textK1\\text{ K}. For aqueous solutions, capprox4.18textJg−1textK−1c \\approx 4.18\\text{ J g}^{-1}\\text{ K}^{-1}.

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Calorimetry is an experimental technique used to measure enthalpy changes. Key assumptions include no heat loss to the surroundings and the density of aqueous solutions being 1.00textgcm−31.00\\text{ g cm}^{-3}.

📐Formulae

q=mcDeltaTq = mc\\Delta T

DeltaH=−fracqn\\Delta H = -\\frac{q}{n}

DeltaT=Tfinal−Tinitial\\Delta T = T_{final} - T_{initial}

n=ctimesVn = c \\times V

n=fracmMn = \\frac{m}{M}

💡Examples

Problem 1:

In a coffee-cup calorimeter, 50.0textcm350.0\\text{ cm}^3 of 1.00textmoldm−3textHCl(aq)1.00\\text{ mol dm}^{-3}\\text{ HCl}(aq) is mixed with 50.0textcm350.0\\text{ cm}^3 of 1.00textmoldm−3textNaOH(aq)1.00\\text{ mol dm}^{-3}\\text{ NaOH}(aq). The temperature rises from 21.0^\\circ\\text{C} to 27.5^\\circ\\text{C}. Calculate the molar enthalpy of neutralization (DeltaHneut\\Delta H_{neut}) in textkJmol−1\\text{kJ mol}^{-1}. Assume the density of the solution is 1.00textgcm−31.00\\text{ g cm}^{-3} and the specific heat capacity is 4.18textJg−1textK−14.18\\text{ J g}^{-1}\\text{ K}^{-1}.

Solution:

  1. Calculate the temperature change DeltaT\\Delta T: beginarrayr27.5−21.0hline6.5endarray\\begin{array}{r} 27.5 \\ -21.0 \\ \\hline 6.5 \\end{array} So, DeltaT=6.5textK\\Delta T = 6.5\\text{ K}.

  2. Calculate the total mass of the solution: m=50.0textg+50.0textg=100.0textgm = 50.0\\text{ g} + 50.0\\text{ g} = 100.0\\text{ g}

  3. Calculate the heat energy qq released: q=mcDeltaTq = mc\\Delta T q=100.0textgtimes4.18textJg−1textK−1times6.5textKq = 100.0\\text{ g} \\times 4.18\\text{ J g}^{-1}\\text{ K}^{-1} \\times 6.5\\text{ K} q=2717textJq = 2717\\text{ J}

  4. Calculate the number of moles of HClHCl or NaOHNaOH (limiting reactant): n=ctimesV=1.00textmoldm−3times0.0500textdm3=0.0500textmoln = c \\times V = 1.00\\text{ mol dm}^{-3} \\times 0.0500\\text{ dm}^3 = 0.0500\\text{ mol}

  5. Calculate the molar enthalpy change DeltaH\\Delta H: DeltaH=−fracqn\\Delta H = -\\frac{q}{n} DeltaH=−frac2717textJ0.0500textmol\\Delta H = -\\frac{2717\\text{ J}}{0.0500\\text{ mol}} DeltaH=−54340textJmol−1\\Delta H = -54340\\text{ J mol}^{-1} DeltaH=−54.3textkJmol−1\\Delta H = -54.3\\text{ kJ mol}^{-1}

Explanation:

The temperature increased, indicating an exothermic reaction, hence the negative sign for DeltaH\\Delta H. We used the total mass of the mixture (100textg100\\text{ g}) for qq, and divided by the moles of the reactant to find the enthalpy change per mole.