Question

The heat of combustion of methane is -880 kJ/mol. If 3.2g of methane is burnt----- of heat is evolved.
a.) 88 kJ
b.) 264 kJ
c.) 176 kJ
d.) 440 kJ

Answer 1

Hint: The heat released when a sample is combusted (heated in presence of air) in its standard state to form stable compounds in their standard states is called heat of combustion.
When we are going to do combustion of carbon in presence of oxygen then the product is carbon dioxide. \[C+{{O}_{2}}\to C{{O}_{2}}\]

Complete step by step answer:
In the question it is given that methane is heated at its combustion value. The chemical reaction of
combustion of methane is as follows.

\[C{{H}_{4}}+2{{O}_{2}}\to C{{O}_{2}}+2{{H}_{2}}O\]
Methane undergoes the combustion process and forms carbon dioxide and water as the products.
Means 16 g of methane undergoes combustion.
In the question it is given that the heat of combustion of methane is -880kJ.
We have to calculate the heat liberated when we burn 3.2 g of methane.
Number of moles of methane burnt = \[\dfrac{3.2}{16}=0.2\] (here molecular weight of methane =16).
Thus 0.2 mole of methane is going to be released when we burn 3.2 g of methane.
Enthalpy of combustion of 1 mole of methane = 880kJ/mol.
Enthalpy of combustion of 0.2 mole of methane = \[880\times 0.2=176kJ\]
Therefore 3.2g of methane is burnt, 176kJ of heat is evolved.
So, the correct answer is “Option C”.

Note: If we are going to heat any compound at higher temperature in the presence of oxygen then it is called combustion and if we are going to heat any compound at higher temperature in the absence of oxygen then it is called pyrolysis.