Question

The enthalpy of combustion of glucose (molecular weight 180 g/mol ) is - 2840 kJ/mol. Then the amount of heat evolved when 0.9 g of glucose is burnt, will be
A ) 14.2 kJ
B ) 1420 kJ
C ) 28.4 kJ
D ) 142 kJ

Answer 1

Hint: Divide mass of glucose with molecular weight to obtain the number of moles of glucose. Multiply number of moles of glucose with the enthalpy of combustion to obtain the amount of heat evolved.

Complete step by step answer:
Enthalpy of combustion is the amount of heat evolved during combustion of one mole of substance. During combustion, glucose reacts with oxygen (from air) to form carbon dioxide gas and water. Write a balanced chemical equation for the combustion of glucose.
\[{{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}{\text{ + 6 }}{{\text{O}}_2}{\text{ }} \to {\text{ 6 C}}{{\text{O}}_2}{\text{ + 6 }}{{\text{H}}_2}{\text{O}}\]
The molecular weight of glucose is 180 g/mol. The mass of glucose is 0.9 g. Divide mass of glucose with its molecular weight to calculate the number of moles of glucose.
\[\dfrac{{0.9{\text{ g}}}}{{180{\text{ g/mol}}}} = 0.005{\text{ mol}}\]
Hence, the number of moles present in the 0.9 g sample of glucose is \[0.005{\text{ }}moles\].
The enthalpy of combustion of glucose is -2840 kJ/mol. Here, negative sign means that energy is given out during combustion reaction. When one mole of glucose undergoes combustion, the energy released is 2840 kJ.
Multiply the number of moles of glucose with its enthalpy of combustion to obtain the amount of heat energy evolved when 0.9 g of glucose undergoes combustion.
\[0.005{\text{ mol }} \times {\text{ 2840 kJ/mol }} = {\text{ 14}}{\text{.2 kJ}}\]
The amount of heat energy evolved when 0.9 g of glucose undergoes combustion is \[14.2{\text{ }}kJ\].

Hence, the option A ) is the correct answer.

Note: It is very important to check proper usage of units. If the enthalpy of combustion is given in terms of kJ/mol , then it is to be multiplied with the number of moles. But if the enthalpy of combustion is given in terms of kJ/g , then it is to be multiplied with the number of grams. If the enthalpy of combustion is given in terms of kJ/mol , and if it is multiplied with the number of grams, then incorrect answers will be obtained.