Example 1
100g solution warms by 5.0∘C. Use c=4.18Jg−1K−1 and neglect other heat transfer. Find reaction heat.
q(solution)=100×4.18×5.0=2090J=2.09 kJ. The reaction released this heat, so q(reaction)=−2.09 kJ.
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Year 11 · SACE Stage 1
Energy changes and calorimetry
Classify energy changes and calculate heat and molar enthalpy.
Method and worked examples
An exothermic reaction transfers energy from the reacting system to the surroundings and has negative ΔH. An endothermic reaction absorbs energy and has positive ΔH. Breaking bonds or separating lattice ions requires energy; forming bonds or hydrating ions releases energy. The net change is the difference between these contributions, so ionic dissolution can be either endothermic or exothermic.
For a simple solution calorimeter use q= mcΔT, with mass m in grams, specific heat capacity c in Jg−1K−1 and temperature change in K or ∘C. If heat exchange with the container and air is neglected, q(reaction)=−q(solution). Divide reaction heat in kJ by reacted moles to obtain ΔH in kJ mol−1. Apply any stated density or heat-capacity assumptions explicitly.
Example 1
Example 2
A reaction releases 3.00 kJ when 0.0500 mol reacts. Find molar enthalpy.
ΔH=0.0500−3.00=−60.0 kJ mol−1. The negative sign denotes an exothermic reaction.
Watch out: A warming solution receives heat; the reaction releases it and has negative ΔH under the stated calorimeter assumptions.