Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![The equilibrium constant, \( K_c \), for the following reaction is \( 7.00 \times 10^{-5} \) at 673 K. Calculate \( K_p \) for this reaction at this temperature.
\[ \text{NH}_4\text{I}(s) \rightleftharpoons \text{NH}_3(g) + \text{HI}(g) \]
\[ K_p = \text{[Box for answer]} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcddd9253-3f40-42a6-9522-fda4ab88f757%2Fa7543fc9-c989-4d8f-9ba4-3cc8b456d6da%2Fd54ccoy9j_processed.png&w=3840&q=75)
Transcribed Image Text:The equilibrium constant, \( K_c \), for the following reaction is \( 7.00 \times 10^{-5} \) at 673 K. Calculate \( K_p \) for this reaction at this temperature.
\[ \text{NH}_4\text{I}(s) \rightleftharpoons \text{NH}_3(g) + \text{HI}(g) \]
\[ K_p = \text{[Box for answer]} \]
![**Equilibrium Constant Calculation**
The equilibrium constant, \( K_p \), for the following reaction is 10.5 at 350 K. Calculate \( K_c \) for this reaction at this temperature.
\[ 2 \text{CH}_2\text{Cl}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{CCl}_4(g) \]
\[ K_c = \underline{\hspace{2cm}} \]
In this problem, you are given the equilibrium constant in terms of pressure (\( K_p \)) and asked to find the equilibrium constant in terms of concentration (\( K_c \)). The reaction involves the conversion of two moles of dichloromethane gas (CH\(_2\)Cl\(_2\)) to one mole of methane gas (CH\(_4\)) and one mole of carbon tetrachloride gas (CCl\(_4\)).
To calculate \( K_c \), use the relation between \( K_p \) and \( K_c \):
\[
K_p = K_c(RT)^{\Delta n}
\]
Where:
- \( R \) is the ideal gas constant (0.0821 L·atm/mol·K).
- \( T \) is the temperature in Kelvin.
- \( \Delta n \) is the change in moles of gas (\( \Delta n = \text{moles of product gases} - \text{moles of reactant gases} = 1 + 1 - 2 = 0 \)).
Since \( \Delta n = 0 \), \( K_p = K_c \) at 350 K, meaning:
\[ K_c = K_p = 10.5 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcddd9253-3f40-42a6-9522-fda4ab88f757%2Fa7543fc9-c989-4d8f-9ba4-3cc8b456d6da%2Fjmc2ong_processed.png&w=3840&q=75)
Transcribed Image Text:**Equilibrium Constant Calculation**
The equilibrium constant, \( K_p \), for the following reaction is 10.5 at 350 K. Calculate \( K_c \) for this reaction at this temperature.
\[ 2 \text{CH}_2\text{Cl}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{CCl}_4(g) \]
\[ K_c = \underline{\hspace{2cm}} \]
In this problem, you are given the equilibrium constant in terms of pressure (\( K_p \)) and asked to find the equilibrium constant in terms of concentration (\( K_c \)). The reaction involves the conversion of two moles of dichloromethane gas (CH\(_2\)Cl\(_2\)) to one mole of methane gas (CH\(_4\)) and one mole of carbon tetrachloride gas (CCl\(_4\)).
To calculate \( K_c \), use the relation between \( K_p \) and \( K_c \):
\[
K_p = K_c(RT)^{\Delta n}
\]
Where:
- \( R \) is the ideal gas constant (0.0821 L·atm/mol·K).
- \( T \) is the temperature in Kelvin.
- \( \Delta n \) is the change in moles of gas (\( \Delta n = \text{moles of product gases} - \text{moles of reactant gases} = 1 + 1 - 2 = 0 \)).
Since \( \Delta n = 0 \), \( K_p = K_c \) at 350 K, meaning:
\[ K_c = K_p = 10.5 \]
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