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25 November, 21:58

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.

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  1. 25 November, 22:06
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    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
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