Questions: Consider these reactions, where M represents a generic metal. 1. 2 M(s) + 6 HCl(aq) -> 2 MCl3(aq) + 3 H2(g) Delta H1 = -793.0 kJ 2. HCl(g) -> HCl(aq) Delta H2 = -74.8 kJ 3. H2(g) + Cl2(g) -> 2 HCl(g) Delta H3 = -1845.0 kJ 4. MCl3(s) -> MCl3(aq) Delta H4 = -257.0 kJ Use the given information to determine the enthalpy of the reaction 2 M(s) + 3 Cl2(g) -> 2 MCl3(s)

Solution Steps

The target reaction is: \[ 2 \mathrm{M}(\mathrm{s}) + 3 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{MCl}_{3}(\mathrm{s}) \]

We need to find the enthalpy change (\(\Delta H\)) for this reaction using the given reactions and their enthalpy changes.

We have the following reactions and their enthalpy changes:

1. \(2 \mathrm{M}(\mathrm{s}) + 6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{MCl}_{3}(\mathrm{aq}) + 3 \mathrm{H}_{2}(\mathrm{g})\), \(\Delta H_{1} = -793.0 \, \mathrm{kJ}\)
2. \(\mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{HCl}(\mathrm{aq})\), \(\Delta H_{2} = -74.8 \, \mathrm{kJ}\)
3. \(\mathrm{H}_{2}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HCl}(\mathrm{g})\), \(\Delta H_{3} = -1845.0 \, \mathrm{kJ}\)
4. \(\mathrm{MCl}_{3}(\mathrm{s}) \rightarrow \mathrm{MCl}_{3}(\mathrm{aq})\), \(\Delta H_{4} = -257.0 \, \mathrm{kJ}\)

To find the enthalpy change for the target reaction, we need to manipulate the given reactions:

• Reverse reaction 4 to get \(\mathrm{MCl}_{3}(\mathrm{aq}) \rightarrow \mathrm{MCl}_{3}(\mathrm{s})\), which changes \(\Delta H_{4}\) to \(+257.0 \, \mathrm{kJ}\).

Combine the reactions to form the target reaction:

1. Start with reaction 1: \[ 2 \mathrm{M}(\mathrm{s}) + 6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{MCl}_{3}(\mathrm{aq}) + 3 \mathrm{H}_{2}(\mathrm{g}) \] \(\Delta H_{1} = -793.0 \, \mathrm{kJ}\)

2. Use reaction 2 to convert \(\mathrm{HCl}(\mathrm{g})\) to \(\mathrm{HCl}(\mathrm{aq})\): \[ 6 \mathrm{HCl}(\mathrm{g}) \longrightarrow 6 \mathrm{HCl}(\mathrm{aq}) \] \(\Delta H_{2} = 6 \times (-74.8) = -448.8 \, \mathrm{kJ}\)

3. Use reaction 3 to form \(\mathrm{HCl}(\mathrm{g})\) from \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\): \[ 3 \mathrm{H}_{2}(\mathrm{g}) + 3 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 6 \mathrm{HCl}(\mathrm{g}) \] \(\Delta H_{3} = 3 \times (-1845.0/2) = -2767.5 \, \mathrm{kJ}\)

4. Reverse reaction 4: \[ 2 \mathrm{MCl}_{3}(\mathrm{aq}) \rightarrow 2 \mathrm{MCl}_{3}(\mathrm{s}) \] \(\Delta H_{4} = 2 \times 257.0 = 514.0 \, \mathrm{kJ}\)

Add the enthalpy changes from the steps above: \[ \Delta H = \Delta H_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} \] \[ \Delta H = -793.0 + (-448.8) + (-2767.5) + 514.0 \] \[ \Delta H = -3495.3 \, \mathrm{kJ} \]

Final Answer