What is the heat of combustion of methane?

Question

A 4.69 g sample of propane, C3H8, is burned in excess oxygen producing carbon dioxide gas and liquid water. The reaction occurs in a calorimeter surrounded by 1.52 kg of water resulting in a temperature change from 23.9°C to 50.1°C. Include the sign in your answer. Assume the heat released by the reaction is completely asborbed by the water.

Answer 1

The energy releases by reaction (this is the heat of combustion) is absorbed by water, so it can be found from the temperature rise of the water:

Energy = mC(delta T)=
E = (1.52kg) x (4.184 kJ/kg/degree) x (50.1-23.9) =
E= -166.62 kJ (negative, b/c it's exothermic)
Combustion is burning, so it always gives off heat (it's exothermic).

The amount of propane that reacted is (4.69g)/(44.09g/mol) = 0.10636 mol.

The heat of combustion is then -166.62kJ/0.10636mol =
-1570 kJ/mol (three sig figs).

Now, if the question was actually asking for the heat of formation of propane (or methane?), then you need to ask a new question.

Answer 2

Methane Heat Of Combustion

Answer 3

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What is the heat of combustion of methane?
A 4.69 g sample of propane, C3H8, is burned in excess oxygen producing carbon dioxide gas and liquid water. The reaction occurs in a calorimeter surrounded by 1.52 kg of water resulting in a temperature change from 23.9°C to 50.1°C. Include the sign in your answer. Assume the heat released by the...

Answer 4

(1.52 kg)(4.186kJ/kg°C)(50.1°C - 23.9°C)/(4.69 g) = 35.527 kJ/g for propane. This says nothing about the heat of combustion of methane (CH4), which has a heat of combustion of about 50.125 kJ/g.

Answer 5

1.52 x 4.18 x (23.9 - 50.1)

This is the enthalpy change in kJ. Notice that I have made the value negative.

To find the molar enthalpy chage, multiply by the Mr of propane (44) and divide by 4.69.