Question

A 35.6 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature rose from 35.0 to 76.0°C and the heat capacity of the calorimeter is 23.3 kJ/°C, what is the value of DH°rxn? The molar mass of ethanol is 46.07 g/mol.             C2H5OH(l) + O2(g) → CO2(g) + H2O(g)                ΔH°rxn = ? (Points : 1)

       -1.24 × 10³ kJ/mol
       +1.24 × 10³ kJ/mol
       -8.09 × 10³ kJ/mol
       -9.55 × 10³ kJ/mol
       +9.55 × 10³ kJ/mol

Answer 1

Q = m c ∆T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g) 35.6 gm
c = specific heat capacity (4.19 for H2O) in J/(g ^oC) 23.3 kJ/°C
∆T = change in temperature = Tfinal - Tinitial in ^oC = 76 -35 = 41 ^oC

q calorimeter : 23.3 kJ/°C x 41°C = 955 kJ

Moles of ethanol burned: 35.6g / 46.07g = 0.773 mol

Combustion is a exothermic reaction, ΔH will be as negative

ΔH°rxn: 955 kJ/ 0.773 mol = -1235 kJ mol-1

Hence answer is -1.24 × 103 kJ/mol