Question

Given,
\[
{\rm{C}}\left( {{\rm{graphite}}} \right) + {{\rm{O}}_2}\left( g \right) \to {\rm{C}}{{\rm{O}}_2}\left( g \right);{\Delta _{\rm{r}}}{\rm{H}}^\circ = - 393.5\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\\
{{\rm{H}}_2}\left( g \right) + \dfrac{1}{2}{{\rm{O}}_2}\left( g \right) \to {{\rm{H}}_2}{\rm{O}}\left( l \right);{\Delta _{\rm{r}}}{\rm{H}}^\circ = - 285.8\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\\
{\rm{C}}{{\rm{O}}_2}\left( g \right) + 2{{\rm{H}}_2}{\rm{O}}\left( l \right) \to {\rm{C}}{{\rm{H}}_4}\left( g \right) + 2{{\rm{O}}_2}\left( g \right){\Delta _{\rm{r}}}{\rm{H}}^\circ = + 890.3\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}
\]
Based on the above thermochemical equations, the value of \[{\Delta _{\rm{r}}}{\rm{H}}^\circ \] at ${\rm{298}}\;{\rm{K}}$ for the reaction
\[{\rm{C}}\left( {{\rm{graphite}}} \right) + 2{{\rm{H}}_2}\left( g \right) \to {\rm{C}}{{\rm{H}}_4}\left( g \right)\] will be:
A. \[ + 144.0\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\]
B. \[ - 144.0\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\]
C. \[ - 74.8\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\]
D. \[ + 78.8\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\]

Answer 1

Hint: We can use the Hess’s law of constant heat summation here while using the given thermochemical values.
We can define the enthalpy of reaction as the enthalpy change that accompanies a given chemical reaction. We can represent the enthalpy of reaction with ${\Delta _{\rm{r}}}{\rm{H}}$ .

Complete step answer
As per the question, our reaction to be considered is:
\[{\rm{C}}\left( {{\rm{graphite}}} \right) + 2{{\rm{H}}_2}\left( g \right) \to {\rm{C}}{{\rm{H}}_4}\left( g \right)\]
Let’s express the above reaction by using the given reactions as follows:
We will take the first reaction as it is \[{\rm{C}}\left( {{\rm{graphite}}} \right) + {{\rm{O}}_2}\left( g \right) \to {\rm{C}}{{\rm{O}}_2}\left( g \right);{\Delta _{\rm{r}}}{\rm{H}}^\circ = - 393.5\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\] because we need graphite to be our reactant.
We will multiply \[{{\rm{H}}_2}\left( g \right) + \dfrac{1}{2}{{\rm{O}}_2}\left( g \right) \to {{\rm{H}}_2}{\rm{O}}\left( l \right);{\Delta _{\rm{r}}}{\rm{H}}^\circ = - 285.8\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\] by two as we require two moles of hydrogen on the reactant side.
So, we will write:
${\rm{2}}{{\rm{H}}_2}\left( g \right) + {{\rm{O}}_2}\left( g \right) \to 2{{\rm{H}}_2}{\rm{O}}\left( l \right);{\Delta _{\rm{r}}}{\rm{H}}^\circ = - \left( {285.8 \times 2} \right)\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}$
Now, we will take the last reaction as it as \[{\rm{C}}{{\rm{O}}_2}\left( g \right) + 2{{\rm{H}}_2}{\rm{O}}\left( l \right) \to {\rm{C}}{{\rm{H}}_4}\left( g \right) + 2{{\rm{O}}_2}\left( g \right);{\Delta _{\rm{r}}}{\rm{H}}^\circ = + 890.3\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\] because we need methane to be our product.
Now, we can use Hess's law of constant heat summation. We can state the Hess’s law of constant heat summation as for those reactions that can occur in multiple steps, we can calculate their reaction enthalpy as the sum of the reaction enthalpies of the individual step reactions given that all reactions are taking place at the same temperature.
Now, let’s write the considered reaction as the sum of the given reactions as follows:
\[
{\rm{C}}\left( {{\rm{graphite}}} \right) + {{\rm{O}}_2}\left( g \right) \to {\rm{C}}{{\rm{O}}_2}\left( g \right);{\Delta _{\rm{r}}}{\rm{H}}^\circ = - 393.5\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\\
{\rm{2}}{{\rm{H}}_2}\left( g \right) + {{\rm{O}}_2}\left( g \right) \to 2{{\rm{H}}_2}{\rm{O}}\left( l \right);{\Delta _{\rm{r}}}{\rm{H}}^\circ = - \left( {285.8 \times 2} \right)\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\\
{\rm{C}}{{\rm{O}}_2}\left( g \right) + 2{{\rm{H}}_2}{\rm{O}}\left( l \right) \to {\rm{C}}{{\rm{H}}_4}\left( g \right) + 2{{\rm{O}}_2}\left( g \right){\Delta _{\rm{r}}}{\rm{H}}^\circ = + 890.3\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\\
{\rm{C}}\left( {{\rm{graphite}}} \right) + 2{{\rm{H}}_2}\left( g \right) \to {\rm{C}}{{\rm{H}}_4}\left( g \right);\;\;\;\;{\Delta _{\rm{r}}}{\rm{H}}^\circ = - 393.5\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}} - \left( {285.8 \times 2} \right)\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}} + 890.3\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}
\]
Finally, we will calculate the reaction enthalpy as follows:
\[
{\Delta _{\rm{r}}}{\rm{H}}^\circ = - 393.5\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}} - \left( {285.8 \times 2} \right)\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}} + 890.3\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\\
 = - {\rm{74}}{\rm{.8}}\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}
\]

Hence, the correct option is C.\[ - 74.8\;{\rm{kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\]

Note:
While using the Hess’s law of constant heat summation, we can also use a reverse reaction of given one that would result in the reverse of the sign of the reaction enthalpy as we can understand the reverse reaction as multiplying by $ - 1$