Question

g The combustion of 1.877 1.877 g of glucose, C 6 H 12 O 6 ( s ) C6H12O6(s), in a bomb calorimeter with a heat capacity of 4.30 4.30 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 22.71 22.71 °C to 29.51 29.51 °C. What is the internal energy change, Δ U ΔU, for the combustion of 1.877 1.877 g of glucose?

Answer 1

Answer:

-2.80 × 10³ kJ/mol

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.

Qcal + Qcomb = 0

Qcomb = - Qcal [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ

where,

C: heat capacity of the calorimeter

ΔT: change in the temperature

From [1],

Qcomb = - Qcal = -29.2 kJ

The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:

ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol