IIT/JEE Chemistry Practice #26: Enthalpy of Combustion

Dave explains a problem involving the enthalpy of combustion for glucose. He outlines the combustion reaction, balances the equation, and performs a thermochemical analysis using standard enthalpies of formation. The enthalpy of combustion is calculated per mole and then converted to per gram of glucose, resulting in -16.11 kJ/g.

What are the standard enthalpies of formation for carbon dioxide, liquid water, and solid glucose at 25 degrees Celsius?

• -300, -400, -1300 kJ/mol

• -500, -200, -1200 kJ/mol

• -400, -300, -1200 kJ/mol

• -400, -300, -1300 kJ/mol

What is the balanced chemical equation for the combustion of glucose?

• C6H12O6 + 6O2 -> 5CO2 + 6H2O

• C6H12O6 + 6O2 -> 6CO2 + 6H2O

• C6H12O6 + 5O2 -> 6CO2 + 6H2O

• C6H12O6 + 6O2 -> 6CO2 + 5H2O

How do you calculate the standard enthalpy change for a reaction?

• Sum of enthalpies of products minus reactants

• Sum of enthalpies of reactants minus products

• Sum of enthalpies of reactants and products

• Difference between enthalpies of reactants and products

What is the enthalpy of combustion for glucose per mole?

• -2800 kJ/mol

• -2700 kJ/mol

• -3000 kJ/mol

• -2900 kJ/mol

How is the enthalpy of combustion per gram of glucose calculated?

• Add the molar mass of glucose

• Divide by the molar mass of glucose

• Multiply by the molar mass of glucose

• Subtract the molar mass of glucose