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3 April, 14:51

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at 25.0∘C. After the reaction, the final temperature of the water is 33.2∘C. Calculate the heat of combustion for 1.00mol of octane.

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  1. 3 April, 15:39
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    The heat of combustion for 1.00 mol of octane is - 5485.7 kJ/mol

    Explanation:

    Step 1: Data given

    Mass of octane = 1.00 grams

    Heat capacity of calorimeter = 837 J/°C

    Mass of water = 1200 grams

    Temperature of water = 25.0°C

    Final temperature : 33.2 °C

    Step 2: Calculate heat absorbed by the calorimeter

    q = c*ΔT

    ⇒ with c = the heat capacity of the calorimeter = 837 J/°C

    ⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

    q = 837 * 8.2 = 6863.4 J

    Step 3: Calculate heat absorbed by the water

    q = m*c*ΔT

    ⇒ m = the mass of the water = 1200 grams

    ⇒ c = the specific heat of water = 4.184 J/g°C

    ⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

    q = 1200 * 4.184 * 8.2 = 41170.56 J

    Step 4: Calculate the total heat

    qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

    Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

    Step 5: Calculate moles of octane

    Moles octane = 1.00 gram / 114.23 g/mol

    Moles octane = 0.00875 moles

    Step 6: Calculate heat combustion for 1.00 mol of octane

    ΔH = - 48 kJ / 0.00875 moles

    ΔH = - 5485.7 kJ/mol

    The heat of combustion for 1.00 mol of octane is - 5485.7 kJ/mol
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