# Homework 44


## 10.49,10.67

Q10.49

Calculate ΔE and ΔH for the evaporation of 1 mol of H2O at a constant pressure. The reaction exerts 2468 kJ of heat and does 14 kJ of work.

1) ΔE=

2) ΔH=

Solution

Strategy:

A. Realize ΔEsystem= q + w (where q= heat and w=work), and the system is the material or process within which we are studying the energy changes in and the surrounds are everything else with which the system can exchange energy with. ΔE is a measure of the system's total energy.

B. Realize ΔEuniverse = ΔEsystem + ΔEsurrounding and ΔEuniverse is constant, therefore when added together ΔEsystem and ΔEsurroundings should add up to 0.

C. When energy flows out of a system ΔEsystem <0, and when energy flows into a system ΔEsystem>0

D. Realize at a constant pressure ΔH = qp where qp is heat at a constant pressure. ΔH is defined as the summation of the system's internal energy and the product of its pressure and volume. Since the pressure in this case is constant it is equal to the negative work and cancels out with the work from the energy equation giving us the equation ΔH = qp. q is equal to the heat.

A. Realize that 2468 kJ is the amount of heat the reaction gives off and 14 kJ is the amount of work the reaction does

B. Realize that the reaction is both giving off heat and exerting work therefore the values will be negative

C. Take the values of both heat and work and add them together to get ΔE.

ΔEsystem= (-2468 kJ) + (-14 kJ)

ΔEsystem= -2482 kJ of work

D. Since ΔH = qp, then qp= -2468

### Q10.67

13.7 g of oxygen gas (O2) combusts in a bomb calorimeter. While undergoing combustion, the temperature rises from 44.1 °C to 48.4 °C. Computed in a completely different experiment, the heat capacity of the bomb calorimeter was found to be 6.24 kJ/°C. Calculate the ΔE for the burning of 1 mol of oxygen gas in kJ/mol oxygen.

1) ΔErxn=

Solution

Strategy:

A. Bomb calorimeters are used to measure ΔE, or since it is a constant volume system. Since the combustion is under constant pressure the equation of ΔEsurroundings = q + w, where work is equal to negative the product of pressure and change in volume (w = -PΔV), is as follows ΔEsurr = qcal

B. To calculate the ΔT or change in temperature, subtract the initial temperature from the final temperature.
C. Realize that heat (q) in the bomb calorimeter is equal to the negative heat of the system, but since it's not possible to measure the temperature of the system, we measure the change in temperature of the surroundings. qcal = Ccal • ΔT.

D. To calculate the number of moles present you have to divide the amount in grams present by the molar mass of the compound or element.

E. To find the ΔE for one mole, place the ΔEsurroundings over the number of moles present.

A. ΔEsurr= q

B. The initial temperature is 44.1 °C, the final temperature is 48.4 °C

ΔT= 48.4°C - 44.1 °C

ΔT=4.3 °C

C. Ccal was calculate to be 6.24 kJ/°C and ΔT was calculated to be 4.3 °C. With stoichiometry the °C should cancel out.

qcal = Ccal • ΔT

qcal = 6.24 kJ/°C • 4.3 °C

qcal= 26.83 kJ

-qcal = ΔEsurroundings

D. The molar mass of oxygen gas is 32.0 g. Given that there is 13.7 g of O2 we can find the number of moles by dividing the given amount by the molar mass.

$\frac{13.7 g }{32.0 g}$ = 0.428 moles O2

E. Place ΔEsurroundings (-26.83 kJ) over number of moles present (0.428 moles)

ΔE= $\frac{-26.83 kJ}{0.428 moles}$

ΔE= -62.69 kJ/mol

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