constant volume caliorimetry

Question

A 1.16-g sample of L-ascorbic acid (C₆H₈O₆) is burned in a bomb calorimeter and the temperature increases from 25.10 °C to 28.10 °C. The calorimeter contains 1.04×10³ g of water and the bomb has a heat capacity of 802 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 = fill in the blank 38cc5305e07f02d_1 kJ/mol

Answer 1

∆E = q + w where q is heat and w is work. Since this is a constant volume situation, w = 0 and ∆E = q.

q = mC∆T + Ccal∆T

q = heat = ?

m = mass = 1.04x10³ g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 28.10 - 25.10 = 3.00º

Ccal = calorimeter constant = 802 J/º

q = mC∆T + Ccal∆T

q = ∆E = (1.04x10³ g)(4.184 J/gº)(3.00º) + (802 J/º)(3.00º) = 1.31x10⁴ J + 2406 J = 1.31x10⁴ J + 0.2406x10⁴J

∆E = 1.55x10⁴ J = 15.5 kJ

To express this on a per mole of ascorbic acid, we find moles of ascorbic acid and then divide ∆E by that value

moles ascorbic acid = 1.16 g x 1 mol/176 g = 0.00659 moles

∆E/mol ascorbic acid = 15.5 kJ/0.00659 moles = 2352 kJ/mole = 2350 kJ/mole (3 sig. figs.)