Question
What will be the amount of heat evolved by burning $$10\,L$$ of methane under standard conditions ?
( Given heats of formation of $$C{H_4},C{O_2}$$ and $${H_2}O$$ are $$ - 76.2, - 398.8$$ and $$ - 241.6\,kJ\,mo{l^{ - 1}}$$ respectively. )
A.
$$805.8\,kJ$$
B.
$$398.8\,kJ$$
C.
$$359.7\,kJ$$
D.
$$640.4\,kJ$$
Answer :
$$359.7\,kJ$$
Solution :
$$\eqalign{
& C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O \cr
& \Delta H = {H_P} - {H_R} \cr
& = \Delta H_f^ \circ \left( {C{O_2}} \right) + 2\Delta H_f^ \circ \left( {{H_2}O} \right) - \left[ {\Delta H_f^ \circ \left( {C{H_4}} \right) + 2\Delta H_f^ \circ \left( {{O_2}} \right)} \right] \cr
& = - 398.8 - 2 \times 241.6 - \left( { - 76.2 + 2 \times 0} \right) \cr
& = - 805.8\,kJ\,mo{l^{ - 1}} \cr} $$
$$22.4\,L\,\,{\text{of}}\,\,C{H_4}\left( {1\,mole} \right)$$ $${\text{gives}}\,\,805.8\,kJ$$
$$10\,L\,\,{\text{of}}\,\,C{H_4}\,\,{\text{will give}}\,\frac{{805.8}}{{22.4}} \times 10$$ $$ = 359.73\,kJ$$