Q.
Given:
$C$ (graphite $)+ O _2( g ) \rightarrow CO _2( g ) ; \Delta_{ r } H ^{ o }=-393.5 \quad kJ mol ^{-1}$
$H _2( g )+\frac{1}{2} O _2( g ) \rightarrow H _2 O ( l ) ; \Delta_{ r } H ^0=-285.8 kJ mol ^{-1}$
$CO _2( g )+2 H _2 O ( l ) \rightarrow CH _4( g )+2 O _2( g ) ; \Delta_{ r } H ^0=+890.3 kJ mol ^{-1}$
Based on the above thermochemical equations, the value of $\Delta_{ r } H ^{\circ}$ at
$298 \quad K$ for the reaction
$C$ (graphite $)+2 H _2( g ) \rightarrow CH _4( g )$ will be:
NTA AbhyasNTA Abhyas 2022
Solution:
The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.
The enthalpy of combustion of a substance is defined as the heat energy given out when one mole of a substance burns completely in oxygen.
The formation reaction of methane is
$C\left(graphite\right)+2H_{2}\left(g\right) \rightarrow \left(CH\right)_{4}\left(g\right)$
The heat of combustion data for $C$ (graphite), $H _2( g )$ and $CH _4( g )$
$
\begin{array}{l}
\Delta H _{ f }^{ o }=\sum \Delta H _{\text {Combustion }}(\text { reactants })-\sum \Delta H _{\text {Combustion }} \text { (products) } \\
\Delta H _{ f }^{ o }\left( CH _4\right)=\Delta H _{ C }^{ o }( C \text { (graphite })+2 \times \Delta H _{ C }^{ o }\left( H _2( g )\right)-\Delta H _{ C }^{ o }\left( CH _4( g )\right) \\
\Delta H _{ f }^{ o }\left( CH _4\right)=-393.5-2 \times 285.8+890.3 \\
\Delta H ^{ o } \text { of } CH _4 ( g )=-74.8 kJ / mol
\end{array}
$
The enthalpy of combustion of a substance is defined as the heat energy given out when one mole of a substance burns completely in oxygen.
The formation reaction of methane is
$C\left(graphite\right)+2H_{2}\left(g\right) \rightarrow \left(CH\right)_{4}\left(g\right)$
The heat of combustion data for $C$ (graphite), $H _2( g )$ and $CH _4( g )$
$ \begin{array}{l} \Delta H _{ f }^{ o }=\sum \Delta H _{\text {Combustion }}(\text { reactants })-\sum \Delta H _{\text {Combustion }} \text { (products) } \\ \Delta H _{ f }^{ o }\left( CH _4\right)=\Delta H _{ C }^{ o }( C \text { (graphite })+2 \times \Delta H _{ C }^{ o }\left( H _2( g )\right)-\Delta H _{ C }^{ o }\left( CH _4( g )\right) \\ \Delta H _{ f }^{ o }\left( CH _4\right)=-393.5-2 \times 285.8+890.3 \\ \Delta H ^{ o } \text { of } CH _4 ( g )=-74.8 kJ / mol \end{array} $