Question

When ethyne is passed through a red hot tube, then the formation of benzene takes place:
$$\begin{gathered} ∆{\mathrm{H}}_{{}_{\mathrm{f}\left({\mathrm{C}}_2{\mathrm{H}}_2\right)}}^{°}\left(\mathrm{g}\right)=230{\mathrm{KJmol}}^{-1} \\ ∆{\mathrm{H}}_{{}_{\mathrm{f}\left({\mathrm{C}}_6{\mathrm{H}}_6\right)}}^{°}\left(\mathrm{g}\right)=85{\mathrm{KJmol}}^{-1} \end{gathered}$$
Calculate the standard heat of trimerisation of ethyne to benzene.
${\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{g}\right)$

• A. $205{\mathrm{KJmol}}^{-1}$
• B. $605{\mathrm{KJmol}}^{-1}$
• C. $-605{\mathrm{KJmol}}^{-1}$
• D. $-205{\mathrm{KJmol}}^{-1}$

Answer 1

The correct option is C

$-605{\mathrm{KJmol}}^{-1}$

Explanation for correct option:

(C)$-605{\mathrm{KJmol}}^{-1}$

Step 1: Trimerization reaction:

• The trimerization reaction is $3{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)\underset{873\mathrm{K}}{\overset{\mathrm{Red}\mathrm{hot}\mathrm{Fe}}{\to }}{\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{g}\right)$

Step 2: Standard heat of formation calculation:

• For calculating the standard heat of formation, the formula is

$∆{\mathrm{H}}_{\mathrm{f}}^{°}=∆{\mathrm{H}}_{\mathrm{f}\left(\mathrm{product}\right)}^{°}-∆{\mathrm{H}}_{\mathrm{f}\left(\mathrm{reactant}\right)}^{°}$

Step 3: Substitution of values:

• On substituting the values, we get

$$\begin{gathered} ∆{\mathrm{H}}_{\mathrm{f}}°=∆{\mathrm{H}}_{\mathrm{f}\left({\mathrm{C}}_6{\mathrm{H}}_6\right)}^{°}-∆{\mathrm{H}}_{\mathrm{f}\left({\mathrm{C}}_2{\mathrm{H}}_2\right)}^{°} \\ =85-3\left(230\right)(3-\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{required}\mathrm{to}\mathrm{react}) \\ =85-690 \\ =-605{\mathrm{KJmol}}^{-1} \end{gathered}$$

So, this is the correct option.

Explanation for incorrect option:

Since the correct option is $-605{\mathrm{KJmol}}^{-1}$. Therefore, options (A), (B) & (D) are incorrect.

Therefore, the correct option is (C) $-605{\mathrm{KJmol}}^{-1}$.