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Chemistry · Ch 7 — Thermodynamics

Measurement of ΔU and ΔH using Calorimetry

7.7

Measurement of ΔU and ΔH using Calorimetry

A calorimeter is the instrument used to measure the amount of heat change accompanying a chemical or physical process. It works indirectly: it measures the resulting TEMPERATURE change (which is inversely proportional to the heat change), and then converts that into a heat quantity using C=q/(mΔT)C=q/(m\Delta T) (from 7.18). Calorimetric measurements are made under two distinct conditions, each answering a different question:

(i) at constant volume, giving qVq_V (which, as Case 2 of Section 7.4.1 showed, equals ΔU\Delta U directly), and …

ΔU Measurement: Bomb Calorimeter

Measuring ΔU\Delta U: the bomb calorimeter. For chemical reactions, heat evolved at constant volume is measured using a bomb calorimeter. The inner vessel -- the "bomb" itself -- and its cover are made of strong steel, with the cover fitted tightly using a metal lid and screws, so the sealed volume genuinely cannot change during the reaction.

A weighed amount of the sample sits in a small platinum cup, connected to electrical ignition wires that can strike an arc to instantly kindle combustion. The bomb is closed tightly and pressurised with excess oxygen, then immersed in water inside the calorimeter's inner volume; a stirrer sits in the gap between the calorimeter wall and the bomb, keeping the surrounding water uniformly mixed. Striking the sample electrically starts the reaction.

A known amount of combustible substance burns in the oxygen inside the sealed bomb. The heat evolved is absorbed both by the calorimeter itself and by the water surrounding it, and the resulting temperature rise is read on a sensitive Beckman thermometer. Because the bomb is sealed, its volume cannot change -- so whatever heat is measured this way IS, by definition, the heat of combustion at CONSTANT VOLUME, ΔUC0\Delta U_C^0.

The total heat produced, ΔUC0\Delta U_C^0, is the sum of the heat absorbed by the calorimeter and by the water:

  • Heat absorbed by the calorimeter: q1=kΔTq_1=k\Delta T, where k=mcCck=m_cC_c is the calorimeter constant (mcm_c = mass of the calorimeter, CcC_c = its heat capacity).
  • Heat absorbed by the water: q2=mwCwΔTq_2=m_wC_w\Delta T, where mwm_w is the mass of water and CwC_w is its molar heat capacity (75.29 J K−1mol−175.29\ \text{J K}^{-1}\text{mol}^{-1}).

ΔUC=q1+q2=kΔT+mwCwΔT=(k+mwCw)ΔT\Delta U_C = q_1+q_2 = k\Delta T+m_wC_w\Delta T = (k+m_wC_w)\Delta T

The calorimeter constant kk itself is found by burning a KNOWN mass of a standard reference sample -- benzoic acid, whose heat of combustion is accurately known (−3227 kJ mol−1-3227\ \text{kJ mol}^{-1}) -- and back-calculating kk from the resulting ΔT\Delta T. …

Figure 7.6Bomb calorimeter

What this figure shows. A cutaway diagram: an inner steel 'steel bomb' vessel holds a small crucible with the yellow powdered 'sample' sitting on an 'ignition coil'/'heater', wired via two red 'ignition wires' running up through the lid; an 'oxygen supply' tube and a mercury 'thermometer' with an 'eyepiece' also enter from the top, alongside a rotating 'Stirrer' paddle. The bomb sits immersed in blue 'water' inside an outer vessel, itself wrapped in an 'insulating jacket' w …

ΔH Measurement: Coffee Cup Calorimeter

Measuring ΔH\Delta H: the coffee-cup calorimeter. Heat change at constant (atmospheric) pressure can instead be measured using a much simpler apparatus -- a coffee-cup calorimeter. Instead of the bomb calorimeter's sealed steel vessel, a styrofoam cup is used. Styrofoam makes a good adiabatic wall: it doesn't allow the heat produced by the reaction to transfer out to the surroundings, so essentially all of that heat energy is absorbed by the water sitting inside the cup itself. This method is suitable specifically for reactions where there is no appreciable change in volume (so no meaningful expansion work needs to be accounted for separately).

The temperature change of the water is measured and converted to a heat quantity using

q=mwCwΔTq = m_wC_w\Delta T

where mwm_w is the mass of water present and CwC_w is its molar heat capacity (75.29 J K−1mol−175.29\ \text{J K}^{-1}\text{mol}^{-1}, the same value used for the bomb calorimeter). …

Figure 7.7Coffee cup Calorimeter

What this figure shows. Nested insulated styrofoam cups holding a light-blue 'Reaction mixture', closed by an 'Insulated stopper' through which a black 'Stirrer' rod and a mercury 'Thermometer' both pass down into the …

Misc 7.4Problem 7.4 -- constant-pressure enthalpy of combustion of ethylene

Worked out. C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l) at 300 K, given constant-volume ΔU=−1406\Delta U=-1406 kJ. Δng=np(g)−nr(g)=2−4=−2\Delta n_g=n_p(g)-n_r(g)=2-4=-2. ΔH=ΔU+RTΔng=−1406+(8.314×10−3×300×(−2))=−1406−4.99=−1410.9\Delta H=\Delta U+RT\Delta n_g=-1406+(8.314\times10^{-3}\times300\times(-2))=-1406-4.99=-1410.9 kJ. …

Misc evaluate-yourself-3Evaluate Yourself 3 -- constant-pressure heat of combustion of benzene

Worked out. Book's practice box (no printed solution): C6H6(l)+712O2(g)→6CO2(g)+3H2O(l)C_6H_6(l)+7\tfrac{1}{2}O_2(g)\rightarrow 6CO_2(g)+3H_2O(l), ΔU\Delta U at 25∘25^\circC =−3268.12=-3268.12 kJ; find ΔH\Delta H. Working it through: Δng=6−7.5=−1.5\Delta n_g=6-7.5=-1.5; ΔH=ΔU+RTΔng=−3268.12+(8.314×10−3)(298)(−1.5)=−3268.12−3.715=−3271.84\Delta H=\Delta U+RT\Delta n_g=-3268.12+(8.314\times10^{-3})(298)(-1.5)=-3268.12-3.715=-3271.84 kJ (own solution, not printed in the textbook). …

Applications of Heat of Combustion and Other Enthalpy Terms

Application 1: calculating a heat of formation from combustion data. Because the heat of combustion of many organic compounds can be measured with relative ease (Sections ~7.7.1/~7.7.2), it is frequently used to back out a heat of FORMATION that would be much harder to measure directly. Worked example: finding ΔHf0\Delta H_f^0 of methane from the known combustion enthalpies of H2_2 (−285.8-285.8), C(graphite) (−393.5-393.5), and CH4_4 itself (−890.4-890.4 kJ mol−1^{-1}), remembering that the standard formation enthalpy of every pure element is defined as zero. Write out the target formation equation and the three combustion equations as a thermochemical set:

C(graphite)+2H2(g)→CH4(g),ΔHf0=X kJ mol−1(i)C(\text{graphite})+2H_2(g)\rightarrow CH_4(g), \quad \Delta H_f^0=X\ \text{kJ mol}^{-1}\quad\text{(i)}

H2(g)+12O2(g)→H2O(l),ΔH0=−285.8 kJ mol−1(ii)H_2(g)+\tfrac{1}{2}O_2(g)\rightarrow H_2O(l), \quad \Delta H^0=-285.8\ \text{kJ mol}^{-1}\quad\text{(ii)}

C(graphite)+O2→CO2,ΔH0=−393.5 kJ mol−1(iii)C(\text{graphite})+O_2\rightarrow CO_2, \quad \Delta H^0=-393.5\ \text{kJ mol}^{-1}\quad\text{(iii)}

CH4(g)+2O2→CO2(g)+2H2O(l),ΔH0=−890.4 kJ mol−1(iv)CH_4(g)+2O_2\rightarrow CO_2(g)+2H_2O(l), \quad \Delta H^0=-890.4\ \text{kJ mol}^{-1}\quad\text{(iv)}

Since methane sits on the PRODUCT side of the target equation (i), reaction (iv) must be reversed:

CO2(g)+2H2O(l)→CH4(g)+2O2,ΔH0=+890.4 kJ mol−1(v)CO_2(g)+2H_2O(l)\rightarrow CH_4(g)+2O_2, \quad \Delta H^0=+890.4\ \text{kJ mol}^{-1}\quad\text{(v)}

Combining by Hess's law, (i) == [(ii)×2\times2] ++ (iii) ++ (v):

X=[(−285.8)×2]+[−393.5]+[+890.4]=−571.6−393.5+890.4=−74.7 kJX = [(-285.8)\times2]+[-393.5]+[+890.4] = -571.6-393.5+890.4 = -74.7\ \text{kJ}

So the heat of formation of methane works out to −74.7 kJ mol−1-74.7\ \text{kJ mol}^{-1}.

Application 2: calorific value of food and fuels. The calorific value is defined as the amount of heat produced (in calories or joules) when one gram of a substance is completely burnt. Its SI unit is J kg−1\text{J kg}^{-1}, though in practice it is usually quoted in cal g−1\text{cal g}^{-1}.

The rest of the enthalpy catalogue. Rounding off the list of named enthalpy quantities begun back in Section 7.5.2:

Heat of solution -- the change in enthalpy when one mole of a substance dissolves in a specified quantity of solvent at a given temperature.

Heat of neutralisation -- "the change in enthalpy when one gram equivalent of an acid is completely neutralised by one gram equivalent of a base (or vice versa) in dilute solution." Remarkably, this comes out to almost exactly the SAME value, −57.32-57.32 kJ, for ANY strong acid neutralised by any strong base:

HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l),ΔH=−57.32 kJHCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l),\quad \Delta H=-57.32\ \text{kJ}

HCl(aq)+KOH(aq)→KCl(aq)+H2O(l),ΔH=−57.32 kJHCl(aq)+KOH(aq)\rightarrow KCl(aq)+H_2O(l),\quad \Delta H=-57.32\ \text{kJ}

HNO3(aq)+KOH(aq)→KNO3(aq)+H2O(l),ΔH=−57.32 kJHNO_3(aq)+KOH(aq)\rightarrow KNO_3(aq)+H_2O(l),\quad \Delta H=-57.32\ \text{kJ}

H2SO4(aq)+2KOH(aq)→K2SO4(aq)+2H2O(l),ΔH=−57.32 kJ×2H_2SO_4(aq)+2KOH(aq)\rightarrow K_2SO_4(aq)+2H_2O(l),\quad \Delta H=-57.32\ \text{kJ}\times2

The reason is Arrhenius's theory of acids and bases: strong acids and strong bases both ionise COMPLETELY in aqueous solution, so in every one of the reactions above, what is really happening -- underneath the spectator ions -- is the identical net-ionic reaction

H+(aq)+OH−(aq)→H2O(l),ΔH=−57.32 kJH^+(aq)+OH^-(aq)\rightarrow H_2O(l),\quad \Delta H=-57.32\ \text{kJ}

which is why the measured heat of neutralisation stays constant regardless of which particular strong acid or strong base was used.

Molar heat of fusion -- "the change in enthalpy when one mole of a solid substance is converted into the liquid state at its melting point." Example, ice: H2O(s)→273KH2O(l)H_2O(s)\xrightarrow{273\text{K}}H_2O(l), ΔHfusion=+5.98\Delta H_{fusion}=+5.98 kJ.

Molar heat of vapourisation -- "the change in enthalpy when one mole of liquid is converted into vapour state at its boiling point." Example, water: H2O(l)→373KH2O(v)H_2O(l)\xrightarrow{373\text{K}}H_2O(v), ΔHvap=+40.626\Delta H_{vap}=+40.626 kJ. …

Misc 7.7.3-worked-formationWorked example -- standard enthalpy of formation of methane from combustion data

Worked out. Given ΔHC\Delta H_C of H2_2, C(graphite) and CH4_4 are −285.8-285.8, −393.5-393.5 and −890.4-890.4 kJ mol−1^{-1}: (ii) H2(g)+12O2→H2O(l)H_2(g)+\tfrac{1}{2}O_2\rightarrow H_2O(l), ΔH0=−285.8\Delta H^0=-285.8; (iii) C(graphite)+O2→CO2C(graphite)+O_2\rightarrow CO_2, ΔH0=−393.5\Delta H^0=-393.5; (iv reversed, v) CO2(g)+2H2O(l)→CH4(g)+2O2CO_2(g)+2H_2O(l)\rightarrow CH_4(g)+2O_2, ΔH0=+890.4\Delta H^0=+890.4. Target (i) C(graphite)+2H2(g)→CH4(g)C(graphite)+2H_2(g)\rightarrow CH_4(g) == [(ii)×2\times2] + (iii) + (v) =[(−285.8)×2]+[−393.5]+[+890.4]=−571.6−393.5+890.4=−74.7= [(-285.8)\times2]+[-393.5]+[+890.4]=-571.6-393.5+890.4=-74.7 kJ mol−1^{-1}. …