calorimetry find energy change

Question

A 1.32-g sample of L-ascorbic acid (C₆H₈O₆) is burned in a bomb calorimeter and the temperature increases from 25.20 °C to 28.10 °C. The calorimeter contains 1.18×10³ g of water and the bomb has a heat capacity of 885 J/°C. Based on this experiment, calculate ΔE for the combustion reaction per mole of L-ascorbic acid burned (kJ/mol).

1. C₆H₈O₆(s) + 5 O₂(g) 6 CO₂(g) + 4 H₂O(l)    E =  kJ/mol

Answer 1

At constant volume, the heat given off or absorbed is equal to the change in internal energy (∆E) = q

q = mC∆T

q = ?

m = mass = 1.18x103 g (assuming a density of 1g/ml for water)

C = 4.184 J/g/deg for water

∆T = 28.10 - 25.20 = 2.9 degrees

Since the calorimeter will absorb some of the heat (885 J/deg), this must be included in the equation, so...

q = mC∆T + Ccal = (1.18x103 g)(4.184 J/g/deg)(2.9 deg) + (885 J/deg)(2.9 deg) = 1.43x104 J + 0.257x104 J

q = 1.69x104 J = -16.9 kJ <-- This is the heat from combustion of 1.32 g of ascorbic acid. To find kJ/mole we convert the 1.3 g to moles: 1.3 g x 1 mol/176 g = 0.00739 moles (3 sig. figs).

Final answer = -16.9 kJ/0.00739 moles = -2287 kJ/mole = -2290 kJ/mole (to 3 sig. figs.)

(NOTE: The value is negative since the temperature increased, meaning heat given off = exothermic)