Respuesta :
Answer:
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 = Â 41170.56 J
Step 4: Calculate the total heat
qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J Â = 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
Answer: -5490 kJ
Explanation:
First find the heat of combustion of 1g of octane (C8H18) using the given information, and then use the molar mass to determine the heat of combustion for 1mol of octane.
First recognize that the heat released by the combustion reaction is entirely absorbed by the water and the calorimeter.
0 = qrxn + qwater + qcalorimeter
qrxn = −(qwater + qcalorimeter)
qrxn=−[(1200.g)(4.184Jg∘C)(33.2∘C−25.0∘C)+(837J∘C)(33.2∘C−25.0∘C)]qrxn=−48033.96J
Therefore, the heat of combustion for 1.00g of octane is about −48.034kJg. Now, multiply this value by the molar mass to determine the heat of combustion for 1mol of octane.
qrxn = 114.232g/mol × −48.034kJ/g × 1mol
qrxn ≈ −5487.02kJ
Rounding the answer to 3 significant figures, we find that the molar heat of combustion for 1mol of octane is approximately −5490kJ.
