Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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].
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
