Question

What will be the amount of heat evolved by burning 10 L of methane under standard conditions?
(Given heats of formation of $\mathop {CH}\nolimits_4 ,{\text{ }}\mathop {CO}\nolimits_2 {\text{ }}and{\text{ }}\mathop H\nolimits_2 \mathop O\nolimits_{}$ are$- 76.2,$ $- 398.8,$and $\mathop { - 2416KJmol}\nolimits^{ - 1}$ )
A.$805.8kj$
B.$398.8kj$
C.$359.7kj$
D.$640.4j$

Answer 1

Heat is evolved when a substance is made for combustion. In thermodynamics enthalpy of combustion is the heat released when a substance is burned under standard conditions. Enthalpy of formation is defined as the heat change during the formation of 1 mole of the substance under standard conditions. ( $T = 25^oC$ , pressure = 1 bar). The total enthalpy change in a reaction will be the sum of intermediate steps in reactions.

Complete step by step answer:
Combustion of methane will be:-
$\mathop {CH}\nolimits_4 {\text{ }} + {\text{ }}\mathop {2O}\nolimits_2 \to {\text{ }}\mathop {CO}\nolimits_2 {\text{ }} + {\text{ }}\mathop H\nolimits_2 \mathop O\nolimits_{}$
We take formation of substance from their elemental state
$\mathop {C + H}\nolimits_2 \to {\text{ }}\mathop {CH}\nolimits_4 \;$ (we will reverse the sign of enthalpy of formation of methane because we require methane in reactant side )
$\mathop {C + O}\nolimits_2 \to {\text{ }}\mathop {CO}\nolimits_4 \;$
$\mathop H\nolimits_2 + \mathop O\nolimits_2 \to {\text{ }}\mathop H\nolimits_2 \mathop O\nolimits_{} \;$
$\mathop {CH}\nolimits_4 {\text{ }} + {\text{ }}\mathop {2O}\nolimits_2 \to {\text{ }}\mathop {CO}\nolimits_2 {\text{ }} + {\text{ }}\mathop {2H}\nolimits_2 \mathop O\nolimits_{}$

Where, $\Delta H$ is the total enthalpy change of the reaction. $\mathop H\nolimits_P$:- it is the enthalpy of the products
$\mathop H\nolimits_R$:- it is the enthalpy of the reaction.
And $\mathop {\Delta H}\nolimits_r$ :- is the enthalpy change during the formation of substance. Formation always occurs in most stable and form elemental states like for carbon dioxide the elemental states are C and oxygen.
$\Delta H = \mathop H\nolimits_P - \mathop H\nolimits_R$
$= \mathop {\Delta H}\nolimits_f (\mathop {CO}\nolimits_2 ) + \mathop {2\Delta H}\nolimits_f (\mathop H\nolimits_2 \mathop O\nolimits_{} ) - [\Delta f(\mathop {CH}\nolimits_4 ) + 2\Delta \mathop H\nolimits_f (\mathop {CO}\nolimits_2 )]$
$= - 398.8 - \left( {2 \times 241.6} \right) - \left( { - 76.2 + 2 \times 0} \right)$
$= - 805.8{\text{ }}kJ$
22.4 L of $CH_4$(1mole) gives = 805.8 kJ
10 L of $CH_4$ will give.
$\dfrac{{805.8}}{{22.4}} \times 10 = 359.73{\text{ }}kJ$
Our required answer is c that is 359.7KJ.

Note:
We can find the enthalpy of any target reaction by finding the intermediate steps in reactions and then adding the values of enthalpies of intermediate steps in the reactions. This definition is applicable according to Hess’s law of constant heat summation.