Question

3. Mothballs are composed primarily of the hydrocarbon naphthalene (C,H). When 1.25 g naphthalene is burned in a bomb calorimeter, the temperature rises from 25.25 °C to 34.33 °C. (Heat capacity of calorimeter 5.11 kJ/°C). Calculate: a) The heat of reaction per gram of naphthalene b) The heat of reaction per mole of naphthalene Given the following standard enthalpy of formations: AH°, [C,H,OHO --277.7 kJ/mol]; AH [CHCOH) = 484.5 kJ/mol]; AH [HO) = -285.8 kJ/mol]; AH® [0,(g) - 0 kJ/mol].

Answer 1

Answer-3 molecular mass of naphthalene (cio Ho) (M] = 12@gm/mol. weight of naphthalene (Coro) (m) = 1.25 gm t = 25.25°C t2 = 34.336 Heat capacity of calorimeter (c) = 5111kJ/ °C Dit = tarit, = 34.33- 25:25 Dt = good'c (a) heat of reaction per gram of naphthalene. DE reaction = cхot micioro) = 5.11 KJ X 9:00 € 1125 gm = 46.3988 KJ 125 gm A Erean = 37.11904 KJ 1 ml

D Erean = (6) Heat of reaction per mole of naphthalene on = cхDA x M(Cioro) meciono) = 5.11 KI x 9.08€ x 120gm mol 125 gm. = 5939 0464 KJ/mol 1.25 berean= 4751.23712 KJ/mol ) Hence, Heat of reaction per gram of naphthalene is Derean = 37.11904 KJ Igm and Heat of reaction per mole of naphthalene. is berean = 4751.23712 KJ/mol