09/14 Thermochemistry

A structured guide to energy transfer, enthalpy, calorimetry, Hess’s law, and quantitative reaction-energy calculations in thermochemistry.

Energy Transfer and the First Law

Thermochemistry examines how energy changes accompany chemical reactions and physical processes. A system is the portion of the universe being studied, such as a reacting mixture; the surroundings include everything outside the system. Energy can cross the system boundary as or work, but total energy is conserved.

  • is energy transferred because of a temperature difference.

  • Work is energy transferred when a force causes matter to move. For a reacting gas, the most common type is pressure–volume work.

  • is the total microscopic energy of the system.

Using the chemistry sign convention:

  • q>0q>0: the system absorbs .

  • q<0q<0: the system releases .

  • w>0w>0: work is done on the system.

  • w<0w<0: work is done by the system.

The first law of thermodynamics is:

ΔU=q+w\Delta U=q+w

For pressure–volume work:

w=−PextΔVw=-P_{\text{ext}}\Delta V

During expansion, ΔV>0\Delta V>0, so w<0w<0: the system does work on the surroundings. During compression, ΔV<0\Delta V<0, so w>0w>0.

For example, if a system absorbs 125 J125\ \mathrm{J} of and does 40 J40\ \mathrm{J} of work on the surroundings, then:

ΔU=(+125 J)+(−40 J)=+85 J\Delta U=(+125\ \mathrm{J})+(-40\ \mathrm{J})=+85\ \mathrm{J}

The increases by 85 J85\ \mathrm{J}. Energy, , and work are measured in joules, while chemical reaction energies are commonly reported in kilojoules.

Takeaway: Energy is conserved, and the first law tracks how energy enters or leaves a system as and work.

Temperature, Capacity, and Specific

Temperature describes the average kinetic energy of particles. It is not the same as : is energy in transit, whereas temperature describes the thermal state of matter.

The required to change the temperature of a sample is:

q=mcΔTq=mc\Delta T

Here, mm is mass, cc is specific capacity, and ΔT=Tfinal−Tinitial\Delta T=T_{\text{final}}-T_{\text{initial}}. Specific capacity is the required to raise the temperature of 1 g1\ \mathrm{g} of a substance by 1 ∘C1\ ^\circ\mathrm{C} or 1 K1\ \mathrm{K}. For liquid water, a commonly used value is:

cwater=4.184 J g−1 ∘C−1c_{\text{water}}=4.184\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}

A positive temperature change gives positive qq, meaning that the sample absorbs . A negative temperature change gives negative qq, meaning that the sample releases .

For 75.0 g75.0\ \mathrm{g} of water warming from 20.0 ∘C20.0\ ^\circ\mathrm{C} to 35.0 ∘C35.0\ ^\circ\mathrm{C}:

q=(75.0 g)(4.184 J g−1 ∘C−1)(15.0 ∘C)=4.71×103 J=4.71 kJq=(75.0\ \mathrm{g})(4.184\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}})(15.0\ ^\circ\mathrm{C})=4.71\times10^3\ \mathrm{J}=4.71\ \mathrm{kJ}

The water absorbs 4.71 kJ4.71\ \mathrm{kJ}.

Takeaway: Use q=mcΔTq=mc\Delta T when the mass, specific , and temperature change are known.

and Reaction

is defined as:

H=U+PVH=U+PV

For a process at constant pressure in which the only work is pressure–volume work:

qp=ΔHq_p=\Delta H

This relationship makes particularly useful for reactions carried out in open containers at approximately constant atmospheric pressure.

  • An exothermic process releases , so qp=ΔH<0q_p=\Delta H<0.

  • An endothermic process absorbs , so qp=ΔH>0q_p=\Delta H>0.

A thermochemical equation must contain a balanced chemical equation and the physical states of all substances because the change depends on the amounts and states involved. For example:

CH4(g)+2O2(g)→CO2(g)+2H2O(l)ΔH=−890.3 kJ\mathrm{CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)}\qquad \Delta H=-890.3\ \mathrm{kJ}

The value applies to the reaction exactly as written. Reversing the equation changes the sign of ΔH\Delta H, and multiplying every coefficient by a factor multiplies ΔH\Delta H by the same factor. is an extensive property, so the energy change scales with the amount of material.

Takeaway: At constant pressure, reaction is represented by ΔH\Delta H; its sign identifies whether the process is exothermic or endothermic.

and Measuring Reaction

determines transfer from measured temperature changes. In an insulated experiment, energy released by the reaction is absorbed by the solution and/or the calorimeter:

qrxn+qsurroundings=0q_{\text{rxn}}+q_{\text{surroundings}}=0

Therefore:

qrxn=−qsurroundingsq_{\text{rxn}}=-q_{\text{surroundings}}

Coffee-cup

A coffee-cup calorimeter operates approximately at constant pressure and is commonly used for aqueous reactions. If the calorimeter capacity is negligible:

qrxn=−qsolution=−msolutioncsolutionΔTq_{\text{rxn}}=-q_{\text{solution}}=-m_{\text{solution}}c_{\text{solution}}\Delta T

If the calorimeter absorbs significant , include its capacity:

qrxn=−(mcΔT+CcalΔT)q_{\text{rxn}}=-\left(mc\Delta T+C_{\text{cal}}\Delta T\right)

For a reaction that warms 100.0 g100.0\ \mathrm{g} of solution from 21.5 ∘C21.5\ ^\circ\mathrm{C} to 28.0 ∘C28.0\ ^\circ\mathrm{C}, using c=4.184 J g−1 ∘C−1c=4.184\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}:

qsolution=(100.0)(4.184)(6.5)=2.72 kJq_{\text{solution}}=(100.0)(4.184)(6.5)=2.72\ \mathrm{kJ}

Thus:

qrxn=−2.72 kJq_{\text{rxn}}=-2.72\ \mathrm{kJ}

The reaction is exothermic. To obtain a molar , divide the reaction by the moles of the limiting reactant or by the moles of reaction represented by the balanced equation.

Bomb

A bomb calorimeter operates at constant volume and is useful for combustion reactions. Because ΔV≈0\Delta V\approx0, pressure–volume work is negligible, so the measured is related most directly to :

qv=ΔUq_v=\Delta U

The reaction is calculated from the absorbed by the water and the bomb:

qrxn=−(mwatercwaterΔT+CbombΔT)q_{\text{rxn}}=-\left(m_{\text{water}}c_{\text{water}}\Delta T+C_{\text{bomb}}\Delta T\right)

Takeaway: uses the temperature change of the surroundings to determine the released or absorbed by a reaction.

and Reaction Pathways

follows from the fact that is a state function: its change depends only on the initial and final states, not on the pathway between them.

To combine thermochemical equations:

  1. Reverse an equation when necessary; change the sign of ΔH\Delta H.

  2. Multiply an equation by a factor when necessary; multiply ΔH\Delta H by the same factor.

  3. Add the equations and their changes.

  4. Cancel substances that appear on both sides.

For example, consider:

C(s)+12O2(g)→CO(g)ΔH1=−110.5 kJ\mathrm{C(s)+\tfrac{1}{2}O_2(g)\rightarrow CO(g)}\qquad \Delta H_1=-110.5\ \mathrm{kJ}
CO(g)+12O2(g)→CO2(g)ΔH2=−283.0 kJ\mathrm{CO(g)+\tfrac{1}{2}O_2(g)\rightarrow CO_2(g)}\qquad \Delta H_2=-283.0\ \mathrm{kJ}

Adding the equations cancels CO(g)\mathrm{CO(g)}:

ΔH=ΔH1+ΔH2=(−110.5)+(−283.0)=−393.5 kJ\Delta H=\Delta H_1+\Delta H_2=(-110.5)+(-283.0)=-393.5\ \mathrm{kJ}

Thus:

C(s)+O2(g)→CO2(g)ΔH=−393.5 kJ\mathrm{C(s)+O_2(g)\rightarrow CO_2(g)}\qquad \Delta H=-393.5\ \mathrm{kJ}

Takeaway: Manipulate chemical equations first, then apply the same operations to their changes.

Standard Enthalpies of Formation

The ΔHf∘\Delta H_f^\circ is the change when exactly 1 mol1\ \mathrm{mol} of a compound forms from its constituent elements in their standard states.

The formation reaction must produce exactly one mole of the compound. For carbon dioxide:

C(graphite)+O2(g)→CO2(g)\mathrm{C(graphite)+O_2(g)\rightarrow CO_2(g)}
ΔHf∘[CO2(g)]=−393.5 kJ mol−1\Delta H_f^\circ[\mathrm{CO_2(g)}]=-393.5\ \mathrm{kJ\ mol^{-1}}

The of an element in its most stable standard state is defined as zero. Examples include:

  • ΔHf∘[O2(g)]=0\Delta H_f^\circ[\mathrm{O_2(g)}]=0

  • ΔHf∘[C(graphite)]=0\Delta H_f^\circ[\mathrm{C(graphite)}]=0

  • ΔHf∘[H2(g)]=0\Delta H_f^\circ[\mathrm{H_2(g)}]=0

Standard-state data generally refer to a pressure of 1 bar1\ \mathrm{bar}, and tabulated values are commonly given at 298.15 K298.15\ \mathrm{K} unless another temperature is specified.

Takeaway: Formation values describe formation of exactly one mole from elements in their standard states, with elemental reference values set to zero.

Calculating Reaction from Formation Data

For a balanced reaction, calculate the by subtracting the total formation of the reactants from that of the products:

ΔHrxn∘=∑nΔHf∘(products)−∑nΔHf∘(reactants)\Delta H_{\text{rxn}}^\circ=\sum n\Delta H_f^\circ(\text{products})-\sum n\Delta H_f^\circ(\text{reactants})

The coefficient nn is the stoichiometric coefficient in the balanced equation. Include every substance, even those with a formation of zero.

For methane combustion:

CH4(g)+2O2(g)→CO2(g)+2H2O(l)\mathrm{CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)}

Using ΔHf∘[CH4(g)]=−74.8 kJ mol−1\Delta H_f^\circ[\mathrm{CH_4(g)}]=-74.8\ \mathrm{kJ\ mol^{-1}}, ΔHf∘[CO2(g)]=−393.5 kJ mol−1\Delta H_f^\circ[\mathrm{CO_2(g)}]=-393.5\ \mathrm{kJ\ mol^{-1}}, ΔHf∘[H2O(l)]=−285.8 kJ mol−1\Delta H_f^\circ[\mathrm{H_2O(l)}]=-285.8\ \mathrm{kJ\ mol^{-1}}, and ΔHf∘[O2(g)]=0\Delta H_f^\circ[\mathrm{O_2(g)}]=0:

ΔHrxn∘=[(−393.5)+2(−285.8)]−[(−74.8)+2(0)]\Delta H_{\text{rxn}}^\circ=[(-393.5)+2(-285.8)]-[(-74.8)+2(0)]
ΔHrxn∘=(−965.1)−(−74.8)=−890.3 kJ mol−1\Delta H_{\text{rxn}}^\circ=(-965.1)-(-74.8)=-890.3\ \mathrm{kJ\ mol^{-1}}

Therefore, combustion of 1 mol1\ \mathrm{mol} of methane as written releases 890.3 kJ890.3\ \mathrm{kJ}. For 0.250 mol0.250\ \mathrm{mol} of methane:

q=(0.250 mol)(−890.3 kJ mol−1)=−222.6 kJq=(0.250\ \mathrm{mol})(-890.3\ \mathrm{kJ\ mol^{-1}})=-222.6\ \mathrm{kJ}

The negative sign indicates that the system releases energy.

Takeaway: Balance the equation, multiply each formation by its coefficient, and apply products minus reactants.

A Quantitative Problem-Solving Strategy

A reliable thermochemistry calculation follows a consistent sequence:

  1. Write and balance the chemical equation, including physical states.

  2. Identify the requested quantity: qq, internal-energy change ΔU\Delta U, change ΔH\Delta H, or energy per mole.

  3. Select the appropriate relationship, such as ΔU=q+w\Delta U=q+w, q=mcΔTq=mc\Delta T, qrxn=−qsurroundingsq_{\text{rxn}}=-q_{\text{surroundings}}, , or the formation- equation.

  4. Convert units consistently, such as grams to moles or joules to kilojoules.

  5. Track signs carefully: absorbed by the system is positive, and released by the system is negative.

  6. Apply stoichiometry to scale the energy to the amount reacting.

  7. Check whether the result is chemically reasonable. Combustion and many acid–base neutralizations are usually exothermic, but the sign must ultimately be determined from data.

  8. Report appropriate significant figures and units, such as kJ mol−1\mathrm{kJ\ mol^{-1}} for a molar reaction or kJ\mathrm{kJ} for the reaction as written.

Final checklist: A complete answer should show the balanced equation, the chosen relationship, substitutions with units, the sign interpretation, and the final result at an appropriate precision.