Answer :
In order to solve the problem, we need to determine the enthalpy change (ΔH) for the combustion of magnesium and identify the reactions that released heat and the one that released the most heat per mole of reactant. Here’s a detailed step-by-step solution:
### Step-by-Step Solution to Find ΔH for Magnesium Combustion:
1. List the Given Reactions and Their Enthalpy Changes:
- Reaction 1:
[tex]\[ \text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g) \][/tex]
[tex]\[ \Delta H = -450 \text{ kJ/mol} \][/tex]
- Reaction 2:
[tex]\[ \text{MgO}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2O(l) \][/tex]
[tex]\[ \Delta H = -82 \text{ kJ/mol} \][/tex]
- Reaction 3:
[tex]\[ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2O(l) \][/tex]
[tex]\[ \Delta H = -286 \text{ kJ/mol} \][/tex]
2. Combining the Reactions to Find Magnesium Combustion Reaction:
To find the enthalpy change for magnesium combustion ([tex]\(\text{Mg}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{MgO}(s)\)[/tex]), we use Hess's Law.
- By adding Reaction 1 and Reaction 2:
[tex]\[ \text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g) \][/tex]
[tex]\[ \text{MgO}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2O(l) \][/tex]
- We can reverse Reaction 3 to cancel out [tex]\(\text{H}_2(g)\)[/tex] and [tex]\(\text{H}_2O(l)\)[/tex]:
[tex]\[ \text{H}_2O(l) \rightarrow \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \][/tex]
3. Using Hess's Law:
- By combining the reactions:
[tex]\[ \text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g) \][/tex]
[tex]\[ \text{MgO}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2O(l) \][/tex]
[tex]\[ \text{H}_2O(l) \rightarrow \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \][/tex]
- The overall combination leads to:
[tex]\[ \text{Mg}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{MgO}(s) \][/tex]
4. Calculate ΔH for Magnesium Combustion:
[tex]\[ \Delta H_{\text{combustion}} = \Delta H_1 + \Delta H_2 - \Delta H_3 \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{\text{combustion}} = (-450 \text{ kJ/mol}) + (-82 \text{ kJ/mol}) - (-286 \text{ kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = -450 + -82 + 286 \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = -246 \text{ kJ/mol} \][/tex]
### Identifying Heat Releasing Reactions:
All given reactions have negative enthalpies, which means they all release heat.
- Reaction 1: [tex]\(-450 \text{ kJ/mol}\)[/tex]
- Reaction 2: [tex]\(-82 \text{ kJ/mol}\)[/tex]
- Reaction 3: [tex]\(-286 \text{ kJ/mol}\)[/tex]
Thus, reactions [1, 2, 3] released heat.
### Identifying the Reaction that Released the Most Heat per Mole:
From the enthalpy changes of the reactions:
- Reaction 1: [tex]\(-450 \text{ kJ/mol}\)[/tex]
- Reaction 2: [tex]\(-82 \text{ kJ/mol}\)[/tex]
- Reaction 3: [tex]\(-286 \text{ kJ/mol}\)[/tex]
The reaction that released the most heat per mole is Reaction 1.
### Final Answers:
- Heat Releasing Reactions: [1, 2, 3]
- Reaction that Released the Most Heat: Reaction 1
The completed table and answers are:
[tex]\[ \begin{tabular}{|l|c|} \hline \multicolumn{1}{|c|}{ \text{Reaction Equation} } & \Delta H( \text{kJ/mol} ) \\ \hline \text{Reaction 1:} \ \text{Mg} (s)+2 \text{HCl} (aq) \rightarrow \text{MgCl} _2(aq)+ \text{H} _2(g) & -450 \\ \hline \text{Reaction 2:} \ \text{MgO} (s)+2 \text{HCl} (aq) \rightarrow \text{MgCl} _2(aq)+ \text{H} _2 O (l) & -82 \\ \hline \text{Reaction 3:} \ \text{H} _2(g)+1 / 2 \text{O} _2(g) \rightarrow \text{H} _2 O (l) & -286 \\ \hline \text{Magnesium combustion:} \ \text{Mg} (s)+1 / 2 \text{O} _2(g) \rightarrow \text{MgO} (s) & -246 \\ \hline \hline \end{tabular} \][/tex]
Reactions [tex]\([1, 2, 3]\)[/tex] released heat.
Reaction [tex]\(1\)[/tex] released the most heat per mole of reactant.
### Step-by-Step Solution to Find ΔH for Magnesium Combustion:
1. List the Given Reactions and Their Enthalpy Changes:
- Reaction 1:
[tex]\[ \text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g) \][/tex]
[tex]\[ \Delta H = -450 \text{ kJ/mol} \][/tex]
- Reaction 2:
[tex]\[ \text{MgO}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2O(l) \][/tex]
[tex]\[ \Delta H = -82 \text{ kJ/mol} \][/tex]
- Reaction 3:
[tex]\[ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2O(l) \][/tex]
[tex]\[ \Delta H = -286 \text{ kJ/mol} \][/tex]
2. Combining the Reactions to Find Magnesium Combustion Reaction:
To find the enthalpy change for magnesium combustion ([tex]\(\text{Mg}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{MgO}(s)\)[/tex]), we use Hess's Law.
- By adding Reaction 1 and Reaction 2:
[tex]\[ \text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g) \][/tex]
[tex]\[ \text{MgO}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2O(l) \][/tex]
- We can reverse Reaction 3 to cancel out [tex]\(\text{H}_2(g)\)[/tex] and [tex]\(\text{H}_2O(l)\)[/tex]:
[tex]\[ \text{H}_2O(l) \rightarrow \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \][/tex]
3. Using Hess's Law:
- By combining the reactions:
[tex]\[ \text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g) \][/tex]
[tex]\[ \text{MgO}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2O(l) \][/tex]
[tex]\[ \text{H}_2O(l) \rightarrow \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \][/tex]
- The overall combination leads to:
[tex]\[ \text{Mg}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{MgO}(s) \][/tex]
4. Calculate ΔH for Magnesium Combustion:
[tex]\[ \Delta H_{\text{combustion}} = \Delta H_1 + \Delta H_2 - \Delta H_3 \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{\text{combustion}} = (-450 \text{ kJ/mol}) + (-82 \text{ kJ/mol}) - (-286 \text{ kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = -450 + -82 + 286 \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = -246 \text{ kJ/mol} \][/tex]
### Identifying Heat Releasing Reactions:
All given reactions have negative enthalpies, which means they all release heat.
- Reaction 1: [tex]\(-450 \text{ kJ/mol}\)[/tex]
- Reaction 2: [tex]\(-82 \text{ kJ/mol}\)[/tex]
- Reaction 3: [tex]\(-286 \text{ kJ/mol}\)[/tex]
Thus, reactions [1, 2, 3] released heat.
### Identifying the Reaction that Released the Most Heat per Mole:
From the enthalpy changes of the reactions:
- Reaction 1: [tex]\(-450 \text{ kJ/mol}\)[/tex]
- Reaction 2: [tex]\(-82 \text{ kJ/mol}\)[/tex]
- Reaction 3: [tex]\(-286 \text{ kJ/mol}\)[/tex]
The reaction that released the most heat per mole is Reaction 1.
### Final Answers:
- Heat Releasing Reactions: [1, 2, 3]
- Reaction that Released the Most Heat: Reaction 1
The completed table and answers are:
[tex]\[ \begin{tabular}{|l|c|} \hline \multicolumn{1}{|c|}{ \text{Reaction Equation} } & \Delta H( \text{kJ/mol} ) \\ \hline \text{Reaction 1:} \ \text{Mg} (s)+2 \text{HCl} (aq) \rightarrow \text{MgCl} _2(aq)+ \text{H} _2(g) & -450 \\ \hline \text{Reaction 2:} \ \text{MgO} (s)+2 \text{HCl} (aq) \rightarrow \text{MgCl} _2(aq)+ \text{H} _2 O (l) & -82 \\ \hline \text{Reaction 3:} \ \text{H} _2(g)+1 / 2 \text{O} _2(g) \rightarrow \text{H} _2 O (l) & -286 \\ \hline \text{Magnesium combustion:} \ \text{Mg} (s)+1 / 2 \text{O} _2(g) \rightarrow \text{MgO} (s) & -246 \\ \hline \hline \end{tabular} \][/tex]
Reactions [tex]\([1, 2, 3]\)[/tex] released heat.
Reaction [tex]\(1\)[/tex] released the most heat per mole of reactant.
