# Homework Help: Bomb Calorimetry

1. Feb 1, 2008

### Mitchtwitchita

Can anybody help me out with this problem please. I'm not sure that I'm doing it right.

A 1.841 g sample of butanoic acid, C4H8O2 (88.11 g/mol) was burned in a bomb calorimeter with excess oxygen. The temperature of the calorimeter and the water before combustion was 17.26 degrees C; after combustion the calorimeter and the water had a temperature of 24.30 degrees C.The calorimeter had a heat capacity of 697 J/K, and contained 1.372 kg of water. use these data to calculate the molar heat of combustion (in kJ) of butanoic acid.

Ti = 290.26 K
Tf = 297.30 K
Heat capacity for calorimeter = 697/1000 = 0.697 kJ/K
Heat capacity for water = 1372 g X 4.184 J/g * degrees K = 5740.448 J/g * degrees K/1000 = 5.74 kJ/g * degrees K
s = heat capacity of calorimeter + heat capacity of water = 0.697 kJ/K + 5.74 kJ/g * degrees K = 6.44 kJ/g * K

q = msdeltaT
=(1.841 g)(6.44 kJ/g * K)(297.30 K - 290.26 K)
=83.47 kJ

Since qcal + qrxn = 0,
qcal = -qrxn

therefore, qrxn = -83.47 kJ

-83.47 kJ X 88.11 g butanoic acid/1 mol butanoic acid = -7.35 x 10^3 kJ/mol

can anybody confirm these results?