Chemistry
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ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![### Calculating \( K_p \) from \( K_c \)
The equilibrium constant, \( K_c \), for the following reaction is \( 5.10 \times 10^{-6} \) at 548 K. Calculate \( K_p \) for this reaction at this temperature.
\[ \text{NH}_4\text{Cl}(s) \rightleftharpoons \text{NH}_3(g) + \text{HCl}(g) \]
#### Provided Data:
- \( K_c = 5.10 \times 10^{-6} \)
- Temperature, \( T = 548 \) K
#### Formula to convert \( K_c \) to \( K_p \):
\[ K_p = K_c(RT)^{\Delta n} \]
Where:
- \( R = 0.0821 \ \text{L·atm·K}^{-1}\text{·mol}^{-1} \) (Ideal gas constant)
- \( \Delta n \) is the change in the number of moles of gas (products - reactants)
For the given reaction:
\[ \Delta n = (1 + 1) - (0) = 2 \]
### Calculation:
1. Substitute the known values into the formula:
\[ K_p = K_c \times (R \times T)^{\Delta n} \]
\[ K_p = 5.10 \times 10^{-6} \times (0.0821 \times 548)^2 \]
2. Calculate \( (0.0821 \times 548) \):
\[ (0.0821 \times 548) = 44.991 \]
3. Raise this result to the power of 2:
\[ (44.991)^2 = 2024.91 \]
4. Multiply by \( K_c \):
\[ K_p = 5.10 \times 10^{-6} \times 2024.91 = 1.03 \times 10^{-2} \]
### Final Answer:
\[ K_p = 1.03 \times 10^{-2} \]](https://content.bartleby.com/qna-images/question/9cce6311-c94c-4650-90f7-705d2febbb40/570234b5-2e75-4145-b508-806bdc8ad6f1/q7xcle_processed.png)
Transcribed Image Text:### Calculating \( K_p \) from \( K_c \)
The equilibrium constant, \( K_c \), for the following reaction is \( 5.10 \times 10^{-6} \) at 548 K. Calculate \( K_p \) for this reaction at this temperature.
\[ \text{NH}_4\text{Cl}(s) \rightleftharpoons \text{NH}_3(g) + \text{HCl}(g) \]
#### Provided Data:
- \( K_c = 5.10 \times 10^{-6} \)
- Temperature, \( T = 548 \) K
#### Formula to convert \( K_c \) to \( K_p \):
\[ K_p = K_c(RT)^{\Delta n} \]
Where:
- \( R = 0.0821 \ \text{L·atm·K}^{-1}\text{·mol}^{-1} \) (Ideal gas constant)
- \( \Delta n \) is the change in the number of moles of gas (products - reactants)
For the given reaction:
\[ \Delta n = (1 + 1) - (0) = 2 \]
### Calculation:
1. Substitute the known values into the formula:
\[ K_p = K_c \times (R \times T)^{\Delta n} \]
\[ K_p = 5.10 \times 10^{-6} \times (0.0821 \times 548)^2 \]
2. Calculate \( (0.0821 \times 548) \):
\[ (0.0821 \times 548) = 44.991 \]
3. Raise this result to the power of 2:
\[ (44.991)^2 = 2024.91 \]
4. Multiply by \( K_c \):
\[ K_p = 5.10 \times 10^{-6} \times 2024.91 = 1.03 \times 10^{-2} \]
### Final Answer:
\[ K_p = 1.03 \times 10^{-2} \]
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