Question

The enthalpy of combustion of methane graphite and dihydrogen at $298K$ are $-890.3$ kJ $mo{l}^{-1}$ and $-285.8$ kJ $mo{l}^{-1}$ respectively. Enthalpy of formation of $C{H}_4\left(g\right)$ will be:

• A. $-78.8$ $kJ$ $mo{l}^{-1}$
• B. $-52.27$ $kJ$ $mo{l}^{-1}$
• C. $+74.8$ $kJ$ $mo{l}^{-1}$
• D. $+52.26$ $kJ$ $mo{l}^{-1}$

Answer 1

Answer: (A)

$-78.8$ $kJ$ $mo{l}^{-1}$

The combustion reactions are shown below.
$C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)\Delta H=-890.3kJ/mol$......(1)

$C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)\Delta H=-393.5kJ/mol$......(2)

$H_2\left(g\right)+0.5O_2\left(g\right)\to H_2\left(l\right)\Delta H=-285.8kJ/mol$......(3)
The equation for the formation of methane is
$C\left(s\right)+2H_2\left(g\right)\to C{H}_4\left(g\right)\Delta H=?$......(4)
Multiply equation 3 with 2
$2H_2\left(g\right)+O_2\left(g\right)\to 2H_2\left(l\right)\Delta H=-2\left(285.8\right)kJ/mol$
Subtract equation 1 from equation (3)
$2H_2\left(g\right)+O_2\left(g\right)\to 2H_2\left(l\right)\Delta H=-2\left(285.8\right)kJ/mol$
-[$C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)\Delta H=-890.3kJ/mol$]
--------------------------------------------------------------------
$2H_2\left(g\right)+C{O}_2\left(g\right)+2H_{2}O\left(l\right)\to C{H}_4\left(g\right)+2H_2\left(l\right)+O_2\left(g\right)\Delta H=-2\left(285.8\right)+890.3kJ/mol$..........(5)
add equation 2 to equation (5) to form equation 4.
$2H_2\left(g\right)+C{O}_2\left(g\right)+2H_{2}O\left(l\right)\to C{H}_4\left(g\right)+2H_2\left(l\right)+O_2\left(g\right)\Delta H=-2\left(285.8\right)+890.3kJ/mol$
+[$C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)\Delta H=-393.5kJ/mol$]
----------------------------------------------------
$C\left(s\right)+2H_2\left(g\right)\to C{H}_4\left(g\right)$

$\Delta H=-393.5+2(-285.8)-(-890.3)=-74.8kJ/mol$