Compute the molar enthalpy of combustion of glucose (C6 H12O6) : C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that combustion of 0.305 g of glucose caused a raise in temperature of 6.30 °C of a calorimeter with total heat capacity C=755 J/°C. Note that what is given is the heat capacity of the calorimeter, not the specific heat capacity of the substance inside the calorimeter. The molecular weight of glucose is 180.2 amu.

The molar enthalpy of combustion of glucose is - 2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c (calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c (calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 * 6.30 + 755 * 6.30 = 4764.5 J (negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = - 4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = - 2819.3 kJ/moles

The molar enthalpy of combustion of glucose is - 2819.3 kJ/mol