Answer to Question #87474 in General Chemistry for taha

Question

A 0.793-g sample of acetylsalicylic acid (C9H8O4) is burned in a bomb calorimeter and the temperature increases from 25.60 °C to 28.70 °C. The calorimeter contains 1.08×103 g of water and the bomb has a heat capacity of 901 J/°C. Based on this experiment, calculate ΔE for the combustion reaction per mole of acetylsalicylic acid burned (kJ/mol).
C9H8O4(s) + 9 O2(g) 9 CO2(g) + 4 H2O(l)

E = ________kJ/mol

Answer 1

Heat released by the reaction of combustion is equal to heat absorbed by water and calorimeter:

$$Q = Q_{water} +Q_{cal} = cm\Delta T + c_{cal}\times \Delta T = 4.186 \frac{J}{g\times ^\circ C}\times 1.08\times10^3 g \times (28.70 ^\circ C - 25.60 ^\circ C) + 901 \frac{J}{^\circ C}\times (28.70 ^\circ C - 25.60 ^\circ C) =16808J$$

$$\Delta E = \frac{Q}{n}$$

$$n = \frac{m}{M} = \frac{0.793 g}{180.16 \frac{g}{mol}} = 0.0044 mol$$

$$\Delta E = \frac{16808 J}{0.0044 mol} = 3820000 \frac{J}{mol} = 3820 \frac{kJ}{mol}$$