Question

When 0.1523 g of liquid pentane (CH) combusts in a bomb calorimeter, the temperature rises from 23.7C to 29.8 C. What is U for the reaction in kJ/mol pentane? The heat capacity of the bomb calorimeter is 5.23 kJ/C.

Answer 1

U for the reaction is 15,048.58 kJ/mol of pentane.

Step-by-step explanation:

Quantity of heat required = heat capacity of bomb calorimeter × temperature rise = 5.23 kJ/C × (29.8 C - 23.7 C) = 5.23 kJ/C × 6.1 C = 31.903 kJ

Moles of pentane = mass/MW

Mass = 0.1523 g

MW of pentane (C5H12) = 72 g/mol

Moles of pentane = 0.1523/72 = 0.00212 mol

U for the reaction = quantity of heat required ÷ moles of pentane = 31.903 kJ ÷ 0.00212 mol = 15,048.58 kJ/mol