Question

When 0.1523 g of liquid pentane (c_5 ​5 ​​ h_{12} ​12 ​​ ) combusts in a bomb calorimeter, the temperature rises from 23.7^{\circ} ​∘ ​​ c to 29.8 ^{\circ} ​∘ ​​

c. what is \deltaδu_{rxn} ​rxn ​​ for the reaction in kj/mol pentane? the heat capacity of the bomb calorimeter is 5.23 kj/^{\circ} ​∘ ​​
c?

Answer 1

q solution = m x c x ΔT m solution = 0.468 g pentane + 1,000 g H2O = 1,000.468 g solution c solution = c water = 4.184 J/g.°C ΔT = Tf - Ti = 23.65 °C - 20.45 °C = 3.20 °C q solution = 1,000.468 g x 4.184 J/g.°C x 3.20 °C = 13.395 kJ