Answer:
The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol
Explanation:
<u>Step 1:</u> 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
<u> Step 2:</u> 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
<u>Step 3:</u> 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
<u>Step 4</u>: 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.
<u>Step 5</u>: Calculate moles of octane
Moles octane = 1.00 gram / 114.23 g/mol
Moles octane = 0.00875 moles
<u>Step 6:</u> 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
