Question

At 381 K, the following reaction takes place:

$H{g}_{2}C{l}_4\left(g\right)+A{l}_{2}C{l}_6\left(g\right)\to 2HgAlC{l}_5\left(g\right)\Delta H=+10kJ$

(a)Offer an explanation for the very smallfor this reaction for the known structures of the compounds

(b) Explain why the small Hin this reaction is evidence against

as the structure of Hg₂Cl₄(g).

Short answer

(a) In this reaction, the difference is seen in the two Al-Cl bond becoming 2 Hg-Cl bonds. Hence, the difference in energy is between the Al-Cl and Hg-Cl.

(b) The reason this cannot be the product is because it would involve so much of bond breaking and reforming. From the original mercury molecule, two Hg-Cl bonds would have to break and an Hg-Hg bond will be formed.

Step-by-step solution

Enthalpy

Enthalpy is an important term in thermodynamics.The enthalpy is equivalent to the heat content of a particular system. The formula is:

$\Delta H=\Delta U+P\Delta V$

Explanation for the small value of enthalpy change and why the value is evidence against the structure of Hg2Cl4.  

(a)The reason for the low value of enthalpy change is because there is difference in the two compounds based on bond enthalpies. In this reaction, the difference is seen in the two Al-Cl bond becoming 2 Hg-Cl bonds. Hence, the difference in energy is between the Al-Cl and Hg-Cl.

(b) Let us consider the molecule below:

The reason this cannot be the product is because it would involve so much of bond breaking and reforming. From the original mercury molecule, two Hg-Cl bonds would have to break and an Hg-Hg bond will be formed.