Calculating heat of reaction from bomb calorimetry data?

Question

A student runs two experiments with a constant-volume "bomb" calorimeter containing 1500.g of water.

First, a 7.000g tablet of benzoic acid (C6H5CO2H) is put into the "bomb" and burned completely in an excess of oxygen. (Benzoic acid is known to have a heat of combustion of /26.454kJg.)

The temperature of the water is observed to rise from 13.00°C to 39.84°C over a time of 7.5 minutes.

Next, 5.410g of ethylene C2H4 are put into the "bomb" and similarly completely burned in an excess of oxygen. This time the temperature of the water rises from 13.00°C to 53.67°C.

The reaction is: C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(g)

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1. Is this reaction exothermic, endothermic, or neither?

1. If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in the second experiment.

1. Calculate the reaction enthalpy ΔHrxn per mole of C2H4.

Answer 1

The reaction is EXOTHERMIC because the temperature of the water increased telling us that the reaction gave off heat.

To determine how much heat was given off, we must first determine the heat capacity of the calorimeter, C_cal.

Heat of combustion of benzoic acid goes into the water and into the calorimeter.

Heat from combustion of benzoic acid = 7.000 g x 26.454 kJ/g = 185.178 kJ

Heat going into the water = q = mC∆T = (1500 g)(4.184 J/gº)(26.84º) = 168.445 kJ

Difference = 185.178 kJ - 168.445 kJ = 16.733 kJ

C_cal = 16.733 kJ/26.84º = 0.6234 kJ/º

In the second experiment, the temperature of the water/calorimeter went up from 13.00º to 53.67º (40.67º change in temp). This tells us the reaction is EXOTHERMIC and heat is RELEASED by the reaction.

To determine the value of the heat released, we do the following calculation...

q = heat = mC∆T + C_cal∆T

q = (1500 g)(4.184 J/gº)(40.67º) + 0.6234 kJ/º x 40.67º = 255.245 kJ + 25.35 kJ

q = 280.60 kJ = heat released by reaction in the second experiment

∆H_rxn per mole C₂H₄:

moles C₂H₄ used = 5.410 g x 1 mol/28.05 g = 0.1929 moles

∆H_rxn = 280.60 kJ/0.1929 moles = 1455 kJ/mole