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Chemistry · Ch 7 — Thermodynamics

Thermochemical Equations

7.6

Thermochemical Equations

A thermochemical equation is simply a normal, balanced, stoichiometric chemical equation that also carries its enthalpy change alongside it. Six conventions govern how these are written and interpreted:

  1. The coefficients in a balanced thermochemical equation are read as moles of reactants and products involved.
  2. The reaction's enthalpy change, ΔHr\Delta H_r, must always be written with its correct sign AND its unit -- neither can be assumed or omitted.
  3. If the chemical reaction is written in reverse, the value of ΔH\Delta H reverses in SIGN but keeps the same magnitude.
  4. The physical state of every species -- gas, liquid, aqueous, or solid, shown in brackets -- must be specified, because ΔH\Delta H genuinely depends on the physical states of both reactants and products (melting or vaporising a species before or after the reaction changes how much energy is involved).
  5. If the whole thermochemical equation is scaled by multiplying every coefficient by some number, the enthalpy change is multiplied by that same number.
  6. A NEGATIVE ΔHr\Delta H_r signals an exothermic reaction; a POSITIVE ΔHr\Delta H_r signals an endothermic one. Worked illustration of rule (iii): 2H2(g)+O2(g)→2H2O(g),ΔHr0=−967.4 kJ2H_2(g)+O_2(g)\rightarrow2H_2O(g),\qquad \Delta H_r^0=-967.4\ \text{kJ} …

Standard Enthalpy of Reaction from Standard Enthalpy of Formation

Standard enthalpy of reaction from standard enthalpies of formation. The standard enthalpy of a reaction is the enthalpy change when all reactants and products are in their standard states (denoted with the superscript 0^0, as in ΔH0\Delta H^0). Once every substance's own ΔHf0\Delta H_f^0 is known, a reaction's ΔHr0\Delta H_r^0 can be calculated WITHOUT ever running that specific reaction in a calorimeter: it is simply the sum of the products' formation enthalpies minus the sum of the reactants', each weighted by its stoichiometric coefficient. For a general reaction aA+bB→cC+dDaA+bB\rightarrow cC+dD:

ΔHr0=∑ΔHf0(products)−∑ΔHf0(reactants)={c ΔHf0(C)+d ΔHf0(D)}−{a ΔHf0(A)+b ΔHf0(B)}\Delta H_r^0 = \sum \Delta H_f^0(\text{products}) - \sum \Delta H_f^0(\text{reactants}) = \{c\,\Delta H_f^0(C)+d\,\Delta H_f^0(D)\} - \{a\,\Delta H_f^0(A)+b\,\Delta H_f^0(B)\}

Worked example (Problem 7.2). The standard enthalpy change for the combustion of ethanol, C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)C_2H_5OH(l)+3O_2(g)\rightarrow2CO_2(g)+3H_2O(l), given the formation enthalpies of C2H5OH(l)C_2H_5OH(l), CO2(g)CO_2(g) and H2O(l)H_2O(l) are −277-277, −393.5-393.5 and −285.5-285.5 kJ mol−1^{-1} respectively (and O2(g)O_2(g), an element in its standard state, contributes zero):

ΔHr0=[2(−393.5)+3(−285.5)]−[(−277)+0]=[−787−856.5]−[−277]=−1643.5+277=−1366.5 kJ\Delta H_r^0 = [2(-393.5)+3(-285.5)] - [(-277)+0] = [-787-856.5]-[-277] = -1643.5+277 = -1366.5\ \text{kJ} …

Misc 7.2Problem 7.2 -- standard enthalpy of combustion of ethanol

Worked out. C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(l), given ΔHf0\Delta H_f^0: C2H5OH(l)=−277C_2H_5OH(l)=-277, CO2(g)=−393.5CO_2(g)=-393.5, H2O(l)=−285.5H_2O(l)=-285.5 kJ mol−1^{-1} (O2O_2 formation enthalpy =0=0). ΔHr0=[2(−393.5)+3(−285.5)]−[(−277)+0]=[−787−856.5]−[−277]=−1643.5+277=−1366.5\Delta H_r^0=[2(-393.5)+3(-285.5)]-[(-277)+0]=[-787-856.5]-[-277]=-1643.5+277=-1366.5 kJ. …

Misc evaluate-yourself-1Evaluate Yourself 1 -- standard enthalpy of the water-gas-shift reaction

Worked out. Book's practice box (no printed solution): calculate ΔHf0\Delta H_f^0 for CO2(g)+H2(g)→CO(g)+H2O(g)CO_2(g)+H_2(g)\rightarrow CO(g)+H_2O(g), given ΔHf0\Delta H_f^0 of CO2(g)CO_2(g), CO(g)CO(g) and H2O(g)H_2O(g) are −393.5-393.5, −111.31-111.31 and −242-242 kJ mol−1^{-1}. Working it through: ΔHr0=[(−111.31)+(−242)]−[(−393.5)]=−353.31+393.5=+40.19\Delta H_r^0=[(-111.31)+(-242)]-[(-393.5)]=-353.31+393.5=+40.19 kJ mol−1^{-1} (own solution, not printed in the textbook). …

Heat of Combustion

Heat (enthalpy) of combustion, ΔHC\Delta H_C, is defined as "the change in enthalpy of a system when one mole of the substance is completely burnt in excess air or oxygen." Because combustion always releases energy, ΔHC\Delta H_C is ALWAYS negative -- there is no such thing as an endothermic combustion.

Worked examples: methane's heat of combustion,

CH4(g)+2O2(g)→CO2(g)+2H2O(l),ΔHC=−87.78 kJ mol−1CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l),\qquad \Delta H_C=-87.78\ \text{kJ mol}^{-1}

and the combustion of carbon,

C(s)+O2(g)→CO2(g),ΔHC=−394.55 kJ mol−1C(s)+O_2(g)\rightarrow CO_2(g),\qquad \Delta H_C=-394.55\ \text{kJ mol}^{-1} …

Molar Heat Capacities (Cp and Cv)

Molar heat capacity. When heat qq is supplied to a system, the constituent molecules absorb it as extra kinetic energy, raising the system's temperature from T1T_1 to T2T_2. This temperature rise is directly proportional to the heat absorbed and inversely proportional to the mass of substance present:

q∝mΔT⇒q=cmΔT⇒c=qmΔT(7.18)q \propto m\Delta T \qquad\Rightarrow\qquad q=cm\Delta T \qquad\Rightarrow\qquad c=\frac{q}{m\Delta T}\qquad(7.18)

The proportionality constant cc here is the heat capacity. When m=1m=1 kg and (T2−T1)=1(T_2-T_1)=1 K, this is called the specific heat capacity: the heat absorbed by one kilogram of a substance to raise its temperature by one kelvin at a specified temperature. When instead 1 MOLE of substance is used, the same quantity is the molar heat capacity cmc_m: the heat absorbed by one mole of substance to raise its temperature by 1 kelvin. Its SI unit is J K−1mol−1\text{J K}^{-1}\text{mol}^{-1}.

Molar heat capacity can be measured under two different conditions, giving two distinct values: at constant volume (CVC_V) or at constant pressure (CPC_P). Starting from the first law, ΔU=q+w=q−P dV\Delta U=q+w=q-P\,dV, so q=ΔU+P dVq=\Delta U+P\,dV (7.19); differentiating with respect to temperature at CONSTANT volume (dV=0dV=0):

(∂q∂T)V=(∂U∂T)V=CV(7.20)\left(\frac{\partial q}{\partial T}\right)_V = \left(\frac{\partial U}{\partial T}\right)_V = C_V \qquad (7.20)

so CVC_V is the rate of change of internal energy with temperature at constant volume. Similarly, the molar heat capacity at constant pressure is defined as the rate of change of enthalpy with temperature at constant pressure:

CP=(∂H∂T)P(7.21)C_P = \left(\frac{\partial H}{\partial T}\right)_P \qquad (7.21)

Relation between CPC_P and CVC_V for an ideal gas. Starting from H=U+PVH=U+PV (7.8), and for 1 mole of ideal gas PV=nRTPV=nRT (7.22), so H=U+nRTH=U+nRT (7.23). Differentiating with respect to TT:

∂H∂T=∂U∂T+nR⇒CP=CV+nR⇒CP−CV=nR(7.24)\frac{\partial H}{\partial T} = \frac{\partial U}{\partial T} + nR \qquad\Rightarrow\qquad C_P = C_V + nR \qquad\Rightarrow\qquad C_P-C_V = nR \qquad(7.24)

(for one mole, simply CP−CV=RC_P-C_V=R). Physically: at constant pressure the system additionally has to do expansion work against its surroundings as it warms, so it needs MORE heat than at constant volume for the identical temperature rise -- which is exactly why CPC_P is always greater than CVC_V.

Calculating ΔU\Delta U and ΔH\Delta H from heat capacities. For one mole of ideal gas, CV=dU/dT⇒dU=CV dTC_V=dU/dT \Rightarrow dU=C_V\,dT; for a finite change, ΔU=CVΔT=CV(T2−T1)\Delta U=C_V\Delta T=C_V(T_2-T_1), and for nn moles, ΔU=nCV(T2−T1)\Delta U=nC_V(T_2-T_1) (7.25). Identically, ΔH=nCP(T2−T1)\Delta H=nC_P(T_2-T_1) (7.26). …

Misc 7.3Problem 7.3 -- $\Delta U$ and $\Delta H$ on heating oxygen

Worked out. 128.0 g of O2_2 (M=32M=32, so n=4n=4 mol) heated from 0∘0^\circC (273 K) to 100∘100^\circC (373 K); CV=21C_V=21, CP=29C_P=29 J mol−1^{-1}K−1^{-1} (difference ≈8\approx8 J mol−1^{-1}K−1≈R^{-1}\approx R). ΔU=nCV(T2−T1)=4×21×100=8400\Delta U=nC_V(T_2-T_1)=4\times21\times100=8400 J =8.4=8.4 kJ. ΔH=nCP(T2−T1)=4×29×100=11600\Delta H=nC_P(T_2-T_1)=4\times29\times100=11600 J =11.6=11.6 kJ. …

Misc evaluate-yourself-2Evaluate Yourself 2 -- heat to warm 180 g of water

Worked out. Book's practice box (no printed solution): heat needed to raise 180 g of water from 25∘25^\circC to 100∘100^\circC, molar heat capacity of water =75.3=75.3 J mol−1^{-1}K−1^{-1}. Working it through: n=180/18=10n=180/18=10 mol; q=nCmΔT=10×75.3×75=56,475q=nC_m\Delta T=10\times75.3\times75=56{,}475 J =56.475=56.475 kJ (own solution, not printed in the textbook). …