Question

$$\begin{gathered} \mathrm{Given}; \\ {\mathrm{C}}_{\left(\mathrm{graphite}\right)}+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right);{∆}_{\mathrm{r}}{\mathrm{H}}^{°}=393.5{\mathrm{kJmol}}^{-1} \\ {\mathrm{H}}_2\left(\mathrm{g}\right)+[1/2]{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right) \\ {\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right)\to {\mathrm{CH}}_4\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right) \\ {∆}_{\mathrm{r}}{\mathrm{H}}^{°}=+890.3{\mathrm{kJmol}}^{-1} \end{gathered}$$

Based on the above thermochemical equations, the value of ${∆}_{\mathrm{r}}{\mathrm{H}}^{\circ }$ at $298\mathrm{K}$ for the reaction ${\mathrm{C}}_{\left(\mathrm{graphite}\right)}+2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{CH}}_4\left(\mathrm{g}\right)$ will be:

• A. ${}^{-}74.8{\mathrm{kJmol}}^{-1}$
• B. ${}^{-}144{\mathrm{kJmol}}^{-1}$
• C. ${}^{+}74.8\mathrm{kJ}$
• D. ${}^{+}144{\mathrm{kJmol}}^{-1}$

Answer 1

The correct option is A

${}^{-}74.8{\mathrm{kJmol}}^{-1}$

Answer : (A)

Step-1

$$\begin{gathered} \mathrm{C}\left(\mathrm{graphite}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right);{∆}_{\mathrm{r}}\mathrm{H}°={}^{-}393.5\dots \dots \dots \dots \left(1\right) \\ {\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right);{∆}_{\mathrm{r}}\mathrm{H}°={}^{-}285.8\dots \dots \dots \dots \dots \left(2\right) \\ {\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right)\to {\mathrm{CH}}_4\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right);{∆}_{\mathrm{r}}\mathrm{H}°={}^{+}890.3\dots \dots \dots \dots \left(3\right) \end{gathered}$$

Step-2

Add reactions (1) and (3)

$$\begin{gathered} \mathrm{C}\left(\mathrm{graphite}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right)\to {\mathrm{CH}}_4\left(\mathrm{g}\right)+{\mathrm{O}}_2\dots \dots \dots \dots \left(4\right) \\ ∆\mathrm{H}=-393.5+890.3=496.8\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Step-3

Multiply reaction (2) with 2

$$\begin{gathered} 2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{i}\right)\dots \dots \dots \dots \left(5\right) \\ ∆\mathrm{H}=-285.8\times 2=-571.6\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Add reactions (4) and (5)

$$\begin{gathered} \mathrm{C}\left(\mathrm{graphite}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{CH}}_4\left(\mathrm{g}\right) \\ ∆\mathrm{H}=496.8-571.6=-\mathbf{74}.\mathbf{8}\mathbf{kJ}\mathbf{/}\mathbf{mol} \end{gathered}$$

Hence option (A) is the correct answer.