CHEM 200/202

Professor Jing GuOffice: EIS-210

All emails are to be sent to:chem200@sdsu.edu

My office hours will be held on zoom on Tuesday from 9:00 to 11:00 am or by appointment (https://SDSU.zoom.us/s/

99415148959)

UPCOMING IMPORTANT DATES

• Lab report due (all parts) due Sep 27th-11:59 pm.

• Owl pre-assignment: limiting reagent Sep. 27th

• Pre-lab: limiting reagent, Sep 27th, 11:59 pm

LECTURE OBJECTIVES• Chapter 5.2

• Describe the technique of calorimetry.

• Calculate and interpret heat and related properties using calorimetry data.

• Chapter 5.3

• Define enthalpy and explain its classification as a state function.

• Calculate the internal energy change of a system.

Coffee Cup Calorimetry: Heat of Solution (dissolution enthalpy)

When a solid dissolves in water, heat may be evolved or absorbed. The heat of dissolution (dissolving) can be determined using a coffee

cup calorimeter.

The enthalpy of solution, enthalpy of dissolution, or heat of solution is the enthalpy change associated with

the dissolution of a substance in a solvent at constant pressure resulting in infinite dilution. The enthalpy of solution is most often

expressed in kJ/mol at constant temperature.

PROBLEMThe salt copper(II) chloride is soluble in water. When 2.68 g CuCl2 is dissolved in 106.00 g water, the

temperature of the solution increases from 25.00 °C to 27.16 °C. Based on this observation, calculate the dissolution enthalpy, ΔdissH, of CuCl2.

Assume that the specific heat capacity of the solution is 4.184 J g-1 °C-1 and that the energy transfer to the calorimeter is negligible.

ΔdissH =  kJ/mol

When 21.6 g of ammonium sulfate ((NH4)2SO4) is dissolved in 120 g of water in a styrofoam calorimeter of negligible heat capacity, the temperature drops from 25.00 to 22.76 °C. Based on this observation, calculate q for the water and

ΔH° for the process, assuming that the heat absorbed by the salt is negligible.(NH4)2SO4(s)  2NH4+(aq) + SO42- (aq)The specific heat of water is 4.184 J °C-1 g-1.

Give the answers in kJ.

qH2O =  kJ

ΔH° =  kJ

BOMB CALORIMETERS• The calorimeters described

up to this point have all been constant pressure calorimeters. They all operate at atmospheric pressure.

• A bomb calorimeter is a sealed system, that is at constant volume, and as such may experience significant pressure increases.

Bomb calorimeters are used to measure the energy produced from reactions that yield a lot of heat and gaseous products.

PROBLEMA bomb calorimeter, or a constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and the energy content of foods.

In an experiment, a 0.4141 g sample of phenanthrene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.146×103 g of water. During the combustion the temperature increases from 25.75 to 28.50 °C. The heat capacity of water is 4.184 J g-1°C-1.

The heat capacity of the calorimeter was determined in a previous experiment to be 935.9 J/°C.

Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of phenanthrene based on these data.

C14H10(s) + (33/2) O2(g) 5 H2O(l) + 14 CO2(g) + Energy

Molar Heat of Combustion = kJ/mol

INTERNAL ENERGY• The sum of all the kinds of energy in a system (kinetic and

potential) is termed the internal energy (U). It is sometimes expressed as E.

• A system can gain or lose internal energy through any number of processes, leading to a change in the internal energy (∆U).

• We can quantify the changes in the internal energy through the heat change (q) and work (w) done on/by the system.

SIGN CONVENTIONSEnergy can be transferred as heat (q) or work (w).

∆U = q + w

q > 0 - system gains heatq < 0 - system loses heat

w > 0 - work done on systemw < 0 - work done by system

q w ∆U

+ + +

+ – Variable

– + Variable

– – –

INTERNAL ENERGY - STATE FUNCTION

• The internal energy (∆U) of a system is what is termed a state function.

• This means that there is no difference to the internal energy of a system if it gains 100 J of energy through heat, or through work.

q w ∆U

+100 J +5 J +105 J

+5 J +100 J +105 J

+25 J +80 J +105 J

–20 J +125 J +105 J

Aspects of the system will be different (e.g. temperature) but the internal energy will

be the same.

ENTHALPY• Enthalpy (H) is used by chemists to describe the

thermodynamics of chemical and physical processes.

• As U, P and V are all state functions, H is a state function as well.

• We can only measure enthalpy changes (∆H), not enthalpy itself.

• If a process takes place at constant pressure (our calorimeters) only U and V can change.

ENTHALPY• If a system expands (∆V>0) the system will

do work (w<0) on the surroundings.

• The sign of the work (w) will always be the opposite of the pressure-volume work (P∆V).

• For a system at constant pressure, the change in heat (qp) is equal to the change in enthalpy (∆H)

ENTHALPY OF REACTIONS• The ∆H value indicates the amount of heat

associated with the reaction, involving that number of moles of reactants.

ENTHALPY OF REACTIONS• The magnitude of the enthalpy change of a

reaction depends on the physical state of the reactants and products.

ENTHALPY CHANGE

• A negative value of ∆H indicates an exothermic process.

• A positive value of ∆H indicates an endothermic process.

QUESTIONWhat is the enthalpy change associated with the

formation of 3.40 moles of HF?H2(g) + F2(g) → 2HF(g) ∆H = -546 kJ

+928 kJ-928 kJ-546 kJ+273 kJ-273 kJ

Answer:ABCDE

PROBLEM• When 1.34 g of zinc reacts with 60.0 mL of 0.750

M HCl, 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc undergoing this reaction.

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)