Question

A 3.600 g sample of glucose is burned in a bimb calorimeter with a heat capacity of 5.250 JK⁻¹. The temperature increases from 296.25 k to 306.93 k. What is the heat reaction of a mole of glucose?

Answer 1

The heat reaction of a mole of glucose can be calculated using bomb calorimetry. We can determine the heat absorbed by the water and the bomb and then find the negative sum of these values to get the heat produced by the reaction. In this case, the heat reaction of a mole of glucose is -48.8 kJ.

Step-by-step explanation:

To calculate the heat reaction of a mole of glucose using bomb calorimetry, we can first calculate the heat absorbed by the water and the bomb. The heat produced by the reaction is equal to the negative of the sum of the heat absorbed by the water and the bomb. Using the given data and the equation Q = mcΔT, we can calculate the heat absorbed by the water and the bomb.

Qwater = (4.184 J/g°C) × (775 g) × (35.6 °C - 23.8 °C) = 38,300 J

Qbomb = 893 J/°C × (35.6 °C - 23.8 °C) = 10,500 J

Therefore, Qrxn = - (Qwater + Qbomb) = - (38,300 J + 10,500 J) = - 48,800 J = - 48.8 kJ.