Answer:
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
