Question

A 0.424 g sample of liquid C5H12 was combusted completely using excess oxygen inside a bomb (constant volume) calorimeter, with the products being carbon dioxide and liquid water. The calorimeter's heat capacity is 4.782 kJ °C-1. If the temperature inside the calorimeter increased from 25.0 °C to 33.4 °C, determine ΔH for this reaction with respect to the system in kJ mol-1 at 298 K. Do not worry about how realistic the final answer is.

Answer 1

Calorimeter's heat capacity is 4.782 kJ °C^-1

Initial temperature = 25 ^oC

FInal temperature = 33.4^oC

Change in temperature (T) = 33.4 - 25 = 8.4^oC.

Heat (q) =  Calorimeter's heat capacity* T = 4.782 kJ °C^-1* 8.4 ^oC = 40.169 kJ

Mass of pentane = 0.424 g

Molar mass of pentane = 72.146 g/mol

Moles of pentane = mass/ molar mass = 0.424 g/ (72.146g/mol) = 0.00588 mol

Thus, H_rxn = q/ Moles = 40.169 kJ/ 0.00588 mol = 6831.46 kJmol^-1

Now, Heat is released so sign should be negative

Hence H_rxn = -6831.46 kJ/mol