1. Tardigrade
  2. Question
  3. Chemistry
  4. The standard enthalpies of formation of H2O(l) and CO2(g) are respectively -286 kJ mol-1 and -394 kJ mol-1. If the standard heat of combustion of CH4(g) is -891 kJ mol-1, then the standard enthalpy of formation of CH4(g) is

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Q. The standard enthalpies of formation of $H_2O(l)$ and $CO_2(g)$ are respectively $-286 \,kJ \,mol^{-1}$ and $-394 \,kJ \,mol^{-1}$. If the standard heat of combustion of $CH_4(g)$ is $-891 \,kJ \,mol^{-1}$, then the standard enthalpy of formation of $CH_4(g)$ is

KEAMKEAM 2020

Solution:

$H _{2}+\frac{1}{2} O _{2} \rightarrow H _{2} O , \Delta_{f} H ^{0}=-286$
$C + O _{2} \rightarrow CO _{2}, \Delta_{f} H ^{0}=-394$
$CH _{4}+2 O _{2} \rightarrow CO _{2}+2 H _{2} O , \Delta_{ C } H ^{0}-891$
$C +2 H _{2} \rightarrow CH _{4} \Delta_{f} H =?$
$-394+(-286 \times 2)+891=-75\, kJ\,mol ^{-1}$