Answer :
To determine for which reactions the enthalpy change of formation, [tex]\(\Delta H_{f}^{\circ}\)[/tex], is equal to the enthalpy change of reaction, [tex]\(\Delta H_{rm}^{\circ}\)[/tex], we need to check which reactions are formation reactions. A formation reaction is defined as the formation of one mole of a compound from its elements in their standard states.
Here are the reactions:
1. [tex]\( \mathrm{Li (s) + \frac{1}{2} Cl_2(g) \rightarrow LiCl (s)} \)[/tex]
2. [tex]\( \mathrm{2 H_2(g) + O_2(g) \rightarrow 2 H_2O (g)} \)[/tex]
3. [tex]\( \mathrm{Li (s) + \frac{1}{2} Cl_2(l) \rightarrow LiCl (s)} \)[/tex]
4. [tex]\( \mathrm{H_2O_2(g) \rightarrow \frac{1}{2} O_2(g) + H_2O (g)} \)[/tex]
5. [tex]\( \mathrm{H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O (g)} \)[/tex]
6. [tex]\( \mathrm{2 Li (s) + Cl_2(g) \rightarrow 2 LiCl (s)} \)[/tex]
Step-by-step analysis of each reaction:
1. [tex]\(\mathrm{Li (s) + \frac{1}{2} Cl_2(g) \rightarrow LiCl (s)}\)[/tex]
- This reaction forms one mole of lithium chloride ([tex]\( \mathrm{LiCl} \)[/tex]) from its elements in their standard states (solid lithium and chlorine gas).
- This is a formation reaction. So, [tex]\(\Delta H_{rm}^{\circ} = \Delta H_{f}^{\circ}\)[/tex] for this reaction.
2. [tex]\(\mathrm{2 H_2(g) + O_2(g) \rightarrow 2 H_2O (g)}\)[/tex]
- This reaction forms two moles of water ([tex]\( \mathrm{H_2O} \)[/tex]), not one mole.
- This is not a standard formation reaction. Therefore, for this reaction, [tex]\(\Delta H_{rm}^{\circ} \neq \Delta H_{f}^{\circ}\)[/tex].
3. [tex]\(\mathrm{Li (s) + \frac{1}{2} Cl_2(l) \rightarrow LiCl (s)}\)[/tex]
- This reaction involves chlorine in its liquid state, not its standard gaseous state.
- This is not strictly a standard formation reaction. Therefore, [tex]\(\Delta H_{rm}^{\circ} \neq \Delta H_{f}^{\circ}\)[/tex].
4. [tex]\(\mathrm{H_2O_2(g) \rightarrow \frac{1}{2} O_2(g) + H_2O (g)}\)[/tex]
- This reaction is a decomposition reaction, not a formation reaction.
- Therefore, [tex]\(\Delta H_{rm}^{\circ} \neq \Delta H_{f}^{\circ}\)[/tex].
5. [tex]\(\mathrm{H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O (g)}\)[/tex]
- This reaction forms one mole of water ([tex]\( \mathrm{H_2O} \)[/tex]) from its elements in their standard states (hydrogen gas and oxygen gas).
- This is a formation reaction. So, [tex]\(\Delta H_{rm}^{\circ} = \Delta H_{f}^{\circ}\)[/tex] for this reaction.
6. [tex]\(\mathrm{2 Li (s) + Cl_2(g) \rightarrow 2 LiCl (s)}\)[/tex]
- This reaction forms two moles of lithium chloride ([tex]\( \mathrm{LiCl} \)[/tex]), not one mole.
- This is not a standard formation reaction. Therefore, [tex]\(\Delta H_{rm}^{\circ} \neq \Delta H_{f}^{\circ}\)[/tex].
From our analysis, the reactions where [tex]\(\Delta H_{rm}^{\circ} = \Delta H_{f}^{\circ}\)[/tex] of the product(s) are:
1. [tex]\( \mathrm{Li (s) + \frac{1}{2} Cl_2(g) \rightarrow LiCl (s)} \)[/tex]
5. [tex]\( \mathrm{H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O (g)} \)[/tex]
Therefore, the correct answers are reactions 1 and 5: [tex]\([1, 5]\)[/tex].
Here are the reactions:
1. [tex]\( \mathrm{Li (s) + \frac{1}{2} Cl_2(g) \rightarrow LiCl (s)} \)[/tex]
2. [tex]\( \mathrm{2 H_2(g) + O_2(g) \rightarrow 2 H_2O (g)} \)[/tex]
3. [tex]\( \mathrm{Li (s) + \frac{1}{2} Cl_2(l) \rightarrow LiCl (s)} \)[/tex]
4. [tex]\( \mathrm{H_2O_2(g) \rightarrow \frac{1}{2} O_2(g) + H_2O (g)} \)[/tex]
5. [tex]\( \mathrm{H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O (g)} \)[/tex]
6. [tex]\( \mathrm{2 Li (s) + Cl_2(g) \rightarrow 2 LiCl (s)} \)[/tex]
Step-by-step analysis of each reaction:
1. [tex]\(\mathrm{Li (s) + \frac{1}{2} Cl_2(g) \rightarrow LiCl (s)}\)[/tex]
- This reaction forms one mole of lithium chloride ([tex]\( \mathrm{LiCl} \)[/tex]) from its elements in their standard states (solid lithium and chlorine gas).
- This is a formation reaction. So, [tex]\(\Delta H_{rm}^{\circ} = \Delta H_{f}^{\circ}\)[/tex] for this reaction.
2. [tex]\(\mathrm{2 H_2(g) + O_2(g) \rightarrow 2 H_2O (g)}\)[/tex]
- This reaction forms two moles of water ([tex]\( \mathrm{H_2O} \)[/tex]), not one mole.
- This is not a standard formation reaction. Therefore, for this reaction, [tex]\(\Delta H_{rm}^{\circ} \neq \Delta H_{f}^{\circ}\)[/tex].
3. [tex]\(\mathrm{Li (s) + \frac{1}{2} Cl_2(l) \rightarrow LiCl (s)}\)[/tex]
- This reaction involves chlorine in its liquid state, not its standard gaseous state.
- This is not strictly a standard formation reaction. Therefore, [tex]\(\Delta H_{rm}^{\circ} \neq \Delta H_{f}^{\circ}\)[/tex].
4. [tex]\(\mathrm{H_2O_2(g) \rightarrow \frac{1}{2} O_2(g) + H_2O (g)}\)[/tex]
- This reaction is a decomposition reaction, not a formation reaction.
- Therefore, [tex]\(\Delta H_{rm}^{\circ} \neq \Delta H_{f}^{\circ}\)[/tex].
5. [tex]\(\mathrm{H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O (g)}\)[/tex]
- This reaction forms one mole of water ([tex]\( \mathrm{H_2O} \)[/tex]) from its elements in their standard states (hydrogen gas and oxygen gas).
- This is a formation reaction. So, [tex]\(\Delta H_{rm}^{\circ} = \Delta H_{f}^{\circ}\)[/tex] for this reaction.
6. [tex]\(\mathrm{2 Li (s) + Cl_2(g) \rightarrow 2 LiCl (s)}\)[/tex]
- This reaction forms two moles of lithium chloride ([tex]\( \mathrm{LiCl} \)[/tex]), not one mole.
- This is not a standard formation reaction. Therefore, [tex]\(\Delta H_{rm}^{\circ} \neq \Delta H_{f}^{\circ}\)[/tex].
From our analysis, the reactions where [tex]\(\Delta H_{rm}^{\circ} = \Delta H_{f}^{\circ}\)[/tex] of the product(s) are:
1. [tex]\( \mathrm{Li (s) + \frac{1}{2} Cl_2(g) \rightarrow LiCl (s)} \)[/tex]
5. [tex]\( \mathrm{H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O (g)} \)[/tex]
Therefore, the correct answers are reactions 1 and 5: [tex]\([1, 5]\)[/tex].
