Free Pearson Edexcel International A-level Chemistry · Unit 2

Enthalpy changes and calorimetry

What you’ll learn

  • Define enthalpy changes with correct amounts and standard states.
  • Calculate molar enthalpy changes from temperature measurements.
  • Explain the direction of common calorimetry errors.

Understand the idea

An enthalpy change describes heat transfer at constant pressure. Exothermic reactions transfer energy to the surroundings and have negative ΔH\Delta H; endothermic reactions have positive ΔH\Delta H. Standard conditions use a pressure of 100 kPa and a specified temperature, commonly 298 K, with each substance in the stated standard state.

Standard formation enthalpy forms one mole of a compound from elements in their standard states. Standard combustion enthalpy burns one mole of a substance completely. Neutralisation enthalpy refers to one mole of water formed in an acid–base reaction, while atomisation enthalpy forms one mole of gaseous atoms. The specified amount is part of each definition.

Calorimetry estimates energy gained by surroundings from their mass, specific heat capacity and temperature rise. The reaction's energy change has the opposite sign. Divide by the reacting amount defined in the equation and convert joules to kilojoules. Heat absorbed by the vessel and heat lost to air can make an exothermic result less negative than the true value; a temperature–time correction helps estimate the rise before heat loss.

Key formulae

q=mcΔTq=mc\Delta T

ΔHreaction=−q1000nin kJ mol−1\Delta H_{\mathrm{reaction}}=-\frac{q}{1000n}\quad\text{in }\mathrm{kJ\,mol^{-1}}

H+(aq)+OH−(aq)→H2O(l)\mathrm{H^+(aq)+OH^-(aq)\rightarrow H_2O(l)}

12Cl2(g)→Cl(g)(one mole of gaseous atoms)\tfrac12\mathrm{Cl_2(g)\rightarrow Cl(g)}\quad\text{(one mole of gaseous atoms)}

Heat gained grows with temperature rise

Conditions. q = mcΔT, m = 100 g and c = 4.18 J g⁻¹ K⁻¹. Constant heat capacity; no phase change, vessel heat capacity or heat loss is included.

Temperature rise / K; Heat gained by solution / kJ.

A 5 K rise means the solution gains 2.09 kJ. For an isolated reaction-plus-solution model the reaction loses the same heat; the positive plotted heat is not a positive reaction enthalpy.

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Worked example

A reaction involving 0.05000.0500 mol of limiting reagent warms 100100 g of solution by 6.006.00 K. Assume c=4.18 J g−1 K−1c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}} and negligible vessel heat capacity. Find the enthalpy change per mole of that reagent.

  1. The solution gains q=100(4.18)(6.00)=2508q=100(4.18)(6.00)=2508 J.
  2. The reaction therefore releases 2.5082.508 kJ.
  3. Divide by 0.05000.0500 mol and retain the negative sign: ΔH=−2.508/0.0500=−50.16 kJ mol−1\Delta H=-2.508/0.0500=-50.16\ \mathrm{kJ\,mol^{-1}}.

Answer: −50.2 kJ mol−1-50.2\ \mathrm{kJ\,mol^{-1}} to three significant figures.

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