Question

The enthalpy change to form 1 mol \({\rm{H}}{{\rm{g}}_{_2}}{\rm{B}}{{\rm{r}}_2}(\;{\rm{s}})\) from the elementat 25⁰C is \( - 206.77\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\), and that of \({\rm{HgBr}}({\rm{g}})\) is \(96.23\;{\rm{kJ}}\;{\rm{mol}}\;{{\rm{m}}^{ - 1}}\). Compute the enthalpy change for the decomposition of to\(2\;{\rm{molHgBr}}({\rm{g}}):\)\({\rm{H}}{{\rm{g}}_{\rm{2}}}{\rm{B}}{{\rm{r}}_2}(s) \to 2{\rm{HgBr}}(g)\)

\(1\;{\rm{molH}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}(s)\)

Short answer

The enthalpy changes for the decomposition of\(1\;{\rm{molH}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}(s)\)to\(2\;{\rm{molHgBr}}({\rm{g}}):\)

\({\rm{H}}{{\rm{g}}_{\rm{2}}}{\rm{B}}{{\rm{r}}_2}(s) \to 2{\rm{HgBr}}(g)\)is,\(\Delta H = 399.22{\rm{kJmo}}{{\rm{l}}^{ - 1}}\)

Step-by-step solution

Given data 

The given reaction is,\({\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}(\;{\rm{s}}) \to 2{\rm{HgBr}}(g)\).

The enthalpy changes to form 1 mol \({\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}(\;{\rm{s}})\) from the element at 25⁰C is \( - 206.77\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)and for \({\rm{HgBr}}({\rm{g}})\) is \(96.23\;{\rm{kJ}}\;{\rm{mol}}\;{{\rm{m}}^{ - 1}}\).

Concept of change in enthalpy of reaction 

Enthalpy change is the standard enthalpy of formation, which has been determined for a vast number of substances. In any general chemical reaction, the reactants undergo chemical changes and combine to give products. It can be represented by the following equation:

\({\mathop{\rm Re}\nolimits} ac\tan t \to \Pr oduct\)

For any such reaction, the change in enthalpy is represented as\(\Delta H\)and is termed as the reaction enthalpy.

The reaction enthalpy is calculated as:

Mathematically,\(\Delta H = \Delta {H_1} + \Delta {H_2}\).

Where,\(\Delta {H_1}\)is change in enthalpy of reactant and \(\Delta {H_2}\) is change in enthalpy of product.

Write reaction for formation of \({\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}\)

The enthalpy changes formation of\({\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}\)(s) from its elements.

\(\Delta {H_f}\left( {{\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}} \right) = - 206.77{\rm{kmo}}{{\rm{l}}^{ - 1}}\)

The enthalpy changes formation of\({\rm{HgBr}}(g)\)from its elements.

\(\Delta {H_f}\left( {{\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}} \right) = 96.23{\rm{kmo}}{{\rm{l}}^{ - 1}}\)

Determine\(\Delta H\)for the reaction written above.

1 The formation of\({\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}\)(s) from its elements:

\(2{\rm{Hg}}(t) + {\rm{B}}{{\rm{r}}_2}(t) \to {\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}(s)\quad \Delta {H_1} = - 206.77{\rm{kJmo}}{{\rm{l}}^{ - 1}}\)

2 The formation of\({\rm{HgBr}}(g)\)from its elements:

\(2{\rm{Hg}}(t) + {\rm{B}}{{\rm{r}}_2}(t) \to 2{\rm{HgBr}}(g)\quad \Delta {H_2} = 2 \times 96.23{\rm{kmolmo}}{{\rm{l}}^{ - 1}}\)

Reverse the reaction 1 and get the final reaction.

\(\begin{aligned}{c}{\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}(s) \to 2{\rm{Hg}}(l) + {\rm{B}}{{\rm{r}}_2}(l)\quad \Delta {H_1} = ( - 1) \times - 206.77{\rm{kJmo}}{{\rm{l}}^{ - 1}}\\2{\rm{Hg}}(l) + {\rm{B}}{{\rm{r}}_2}(l) \to 2{\rm{HgBr}}(g)\quad \Delta {H_2} = 2 \times 96.23{\rm{kJmol}}^{- 1}\end{aligned}\)

Put all above reaction together.

\(\begin{aligned}{}2{\rm{Hg}}(l) + {\rm{B}}{{\rm{r}}_2}(l) \to {\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}(s)\quad \Delta {H_1} = 206.77{\rm{kJmo}}{{\rm{l}}^{ - 1}}\\2{\rm{Hg}}(l) + {\rm{B}}{{\rm{r}}_2}(l) \to 2{\rm{HgBr}}(g)\quad \Delta {H_2} = 192.46{\rm{kmo}}{{\rm{l}}^{ - 1}}\\{\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}(\;{\rm{s}}) + 2{\rm{Hg}}(l) + {\rm{B}}{{\rm{r}}_2}(l) \to 2{\rm{Hg}}(l) + {\rm{B}}{{\rm{r}}_2}(l) + 2{\rm{HgBr}}(g)\end{aligned}\)

Final equation is,\({\rm{H}}{{\rm{g}}_2}{\rm{B}}{{\rm{r}}_2}(s) \to 2{\rm{HgBr}}(g)\).

Calculation of enthalpy change 

For the final enthalpy change add the enthalpy of two molecules.

\(\begin{aligned}{}\Delta H = \Delta {H_1} + \Delta {H_2}\\\Delta H = 206.77 + 192.46\\\Delta H = 399.22{\rm{kJmo}}{{\rm{l}}^{ - 1}}\end{aligned}\)