Answer :
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].
