Question

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.

Answer 1

Answer: -5490 kJ

Step-by-step 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.

Answer 2

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

Step-by-step 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