At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Absolutely, let's walk through the solution step by step.
### Given Data:
1. Total pressure ([tex]\(P_{total}\)[/tex]): [tex]\(1.00 \times 10^5 \, \text{Pa}\)[/tex]
2. Temperature ([tex]\(T\)[/tex]): [tex]\(500 \, \text{K}\)[/tex]
3. Initial fraction of [tex]\(Cl_2(g)\)[/tex]: [tex]\(1.0\)[/tex]
4. Fraction of [tex]\(Cl_2(g)\)[/tex] that has reacted: [tex]\(70\%\)[/tex] or [tex]\(0.70\)[/tex]
### Step-by-Step Solution:
1. Calculate the initial pressure of [tex]\(Cl_2\)[/tex] (Initial [tex]\(P_{Cl_2}\)[/tex]):
Since [tex]\(Cl_2\)[/tex] initially represents 100% of the total pressure:
[tex]\[ P_{Cl_2 \, \text{initial}} = P_{total} \times \text{Initial fraction of } Cl_2 \][/tex]
[tex]\[ P_{Cl_2 \, \text{initial}} = 1.00 \times 10^5 \, \text{Pa} \times 1.0 = 100000 \, \text{Pa} \][/tex]
2. Calculate the pressure of [tex]\(Cl_2\)[/tex] that has reacted (Reacted [tex]\(P_{Cl_2}\)[/tex]):
[tex]\[ P_{Cl_2 \, \text{reacted}} = P_{Cl_2 \, \text{initial}} \times \text{Fraction reacted} \][/tex]
[tex]\[ P_{Cl_2 \, \text{reacted}} = 100000 \, \text{Pa} \times 0.70 = 70000 \, \text{Pa} \][/tex]
3. Calculate the remaining pressure of [tex]\(Cl_2\)[/tex] (Remaining [tex]\(P_{Cl_2}\)[/tex]):
[tex]\[ P_{Cl_2 \, \text{remaining}} = P_{Cl_2 \, \text{initial}} - P_{Cl_2 \, \text{reacted}} \][/tex]
[tex]\[ P_{Cl_2 \, \text{remaining}} = 100000 \, \text{Pa} - 70000 \, \text{Pa} = 30000 \, \text{Pa} \][/tex]
4. Calculate the pressure of [tex]\(O_2\)[/tex] produced (Produced [tex]\(P_{O_2}\)[/tex]):
For every 2 moles of [tex]\(Cl_2\)[/tex] reacted, 1 mole [tex]\(O_2\)[/tex] is produced. Thus, the pressure of [tex]\(O_2\)[/tex] produced is half the reacted [tex]\(Cl_2\)[/tex] pressure:
[tex]\[ P_{O_2 \, \text{produced}} = \frac{P_{Cl_2 \, \text{reacted}}}{2} \][/tex]
[tex]\[ P_{O_2 \, \text{produced}} = \frac{70000 \, \text{Pa}}{2} = 35000 \, \text{Pa} \][/tex]
5. Calculate the equilibrium constant [tex]\(K_p\)[/tex]:
[tex]\[ K_p = \frac{P_{O_2 \, \text{produced}}}{P_{Cl_2 \, \text{remaining}}^2} \][/tex]
Substituting the values:
[tex]\[ K_p = \frac{35000 \, \text{Pa}}{(30000 \, \text{Pa})^2} \][/tex]
[tex]\[ K_p = \frac{35000}{900000000} \, \text{Pa}^{-1} = 3.888888888888889 \times 10^{-5} \, \text{Pa}^{-1} \][/tex]
6. State the units of [tex]\(K_p\)[/tex]:
Given the formula [tex]\(K_p = \frac{P_{O_2}}{P_{Cl_2}^2}\)[/tex], the units will be:
[tex]\[ \text{Units of } K_p = \text{units of } \frac{\text{Pa}}{\text{Pa}^2} = \text{Pa}^{-1} \][/tex]
### Conclusion:
The equilibrium constant [tex]\(K_p\)[/tex] is [tex]\(3.888888888888889 \times 10^{-5}\)[/tex] and the units are [tex]\(\text{Pa}^{-1}\)[/tex].
### Given Data:
1. Total pressure ([tex]\(P_{total}\)[/tex]): [tex]\(1.00 \times 10^5 \, \text{Pa}\)[/tex]
2. Temperature ([tex]\(T\)[/tex]): [tex]\(500 \, \text{K}\)[/tex]
3. Initial fraction of [tex]\(Cl_2(g)\)[/tex]: [tex]\(1.0\)[/tex]
4. Fraction of [tex]\(Cl_2(g)\)[/tex] that has reacted: [tex]\(70\%\)[/tex] or [tex]\(0.70\)[/tex]
### Step-by-Step Solution:
1. Calculate the initial pressure of [tex]\(Cl_2\)[/tex] (Initial [tex]\(P_{Cl_2}\)[/tex]):
Since [tex]\(Cl_2\)[/tex] initially represents 100% of the total pressure:
[tex]\[ P_{Cl_2 \, \text{initial}} = P_{total} \times \text{Initial fraction of } Cl_2 \][/tex]
[tex]\[ P_{Cl_2 \, \text{initial}} = 1.00 \times 10^5 \, \text{Pa} \times 1.0 = 100000 \, \text{Pa} \][/tex]
2. Calculate the pressure of [tex]\(Cl_2\)[/tex] that has reacted (Reacted [tex]\(P_{Cl_2}\)[/tex]):
[tex]\[ P_{Cl_2 \, \text{reacted}} = P_{Cl_2 \, \text{initial}} \times \text{Fraction reacted} \][/tex]
[tex]\[ P_{Cl_2 \, \text{reacted}} = 100000 \, \text{Pa} \times 0.70 = 70000 \, \text{Pa} \][/tex]
3. Calculate the remaining pressure of [tex]\(Cl_2\)[/tex] (Remaining [tex]\(P_{Cl_2}\)[/tex]):
[tex]\[ P_{Cl_2 \, \text{remaining}} = P_{Cl_2 \, \text{initial}} - P_{Cl_2 \, \text{reacted}} \][/tex]
[tex]\[ P_{Cl_2 \, \text{remaining}} = 100000 \, \text{Pa} - 70000 \, \text{Pa} = 30000 \, \text{Pa} \][/tex]
4. Calculate the pressure of [tex]\(O_2\)[/tex] produced (Produced [tex]\(P_{O_2}\)[/tex]):
For every 2 moles of [tex]\(Cl_2\)[/tex] reacted, 1 mole [tex]\(O_2\)[/tex] is produced. Thus, the pressure of [tex]\(O_2\)[/tex] produced is half the reacted [tex]\(Cl_2\)[/tex] pressure:
[tex]\[ P_{O_2 \, \text{produced}} = \frac{P_{Cl_2 \, \text{reacted}}}{2} \][/tex]
[tex]\[ P_{O_2 \, \text{produced}} = \frac{70000 \, \text{Pa}}{2} = 35000 \, \text{Pa} \][/tex]
5. Calculate the equilibrium constant [tex]\(K_p\)[/tex]:
[tex]\[ K_p = \frac{P_{O_2 \, \text{produced}}}{P_{Cl_2 \, \text{remaining}}^2} \][/tex]
Substituting the values:
[tex]\[ K_p = \frac{35000 \, \text{Pa}}{(30000 \, \text{Pa})^2} \][/tex]
[tex]\[ K_p = \frac{35000}{900000000} \, \text{Pa}^{-1} = 3.888888888888889 \times 10^{-5} \, \text{Pa}^{-1} \][/tex]
6. State the units of [tex]\(K_p\)[/tex]:
Given the formula [tex]\(K_p = \frac{P_{O_2}}{P_{Cl_2}^2}\)[/tex], the units will be:
[tex]\[ \text{Units of } K_p = \text{units of } \frac{\text{Pa}}{\text{Pa}^2} = \text{Pa}^{-1} \][/tex]
### Conclusion:
The equilibrium constant [tex]\(K_p\)[/tex] is [tex]\(3.888888888888889 \times 10^{-5}\)[/tex] and the units are [tex]\(\text{Pa}^{-1}\)[/tex].
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
