The equilibrium constant, Kc, for the following reaction is 57.6 at 277 K. 2CH₂Cl2 (9) CH4 (9) + CCl4 (9) When a sufficiently large sample of CH₂Cl₂ (g) is introduced into an evacuated vessel at 277 K, the equilibrium concentration of CC14 (g) is found to be 0.383 M. Calculate the concentration of CH₂Cl2 in the equilibrium mixture. [CH₂ Cl₂] = M
The equilibrium constant, Kc, for the following reaction is 57.6 at 277 K. 2CH₂Cl2 (9) CH4 (9) + CCl4 (9) When a sufficiently large sample of CH₂Cl₂ (g) is introduced into an evacuated vessel at 277 K, the equilibrium concentration of CC14 (g) is found to be 0.383 M. Calculate the concentration of CH₂Cl2 in the equilibrium mixture. [CH₂ Cl₂] = M
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 57.6 at 277 K.
\[ 2 \text{CH}_2\text{Cl}_2 (g) \rightleftharpoons \text{CH}_4 (g) + \text{CCl}_4 (g) \]
When a sufficiently large sample of \(\text{CH}_2\text{Cl}_2 (g)\) is introduced into an evacuated vessel at 277 K, the equilibrium concentration of \(\text{CCl}_4 (g)\) is found to be 0.383 M.
Calculate the concentration of \(\text{CH}_2\text{Cl}_2\) in the equilibrium mixture.
\[
[\text{CH}_2\text{Cl}_2] = \boxed{\,\,} \,\, \text{M}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e7279b6-864f-464c-a490-1592f397f56d%2Fde8005ae-d373-4d28-8f86-a5ea6f675e4d%2F5toy6cr_processed.png&w=3840&q=75)
Transcribed Image Text:The equilibrium constant, \( K_c \), for the following reaction is 57.6 at 277 K.
\[ 2 \text{CH}_2\text{Cl}_2 (g) \rightleftharpoons \text{CH}_4 (g) + \text{CCl}_4 (g) \]
When a sufficiently large sample of \(\text{CH}_2\text{Cl}_2 (g)\) is introduced into an evacuated vessel at 277 K, the equilibrium concentration of \(\text{CCl}_4 (g)\) is found to be 0.383 M.
Calculate the concentration of \(\text{CH}_2\text{Cl}_2\) in the equilibrium mixture.
\[
[\text{CH}_2\text{Cl}_2] = \boxed{\,\,} \,\, \text{M}
\]
![**Equilibrium Constant Calculation**
The equilibrium constant, \( K_c \), for the following reaction is \(1.80 \times 10^{-4}\) at 298 K.
\[ \text{NH}_4\text{HS} (s) \rightleftharpoons \text{NH}_3 (g) + \text{H}_2\text{S} (g) \]
**Problem Statement:**
Calculate the equilibrium concentration of \(\text{H}_2\text{S}\) when 0.525 moles of \(\text{NH}_4\text{HS} (s)\) are introduced into a 1.00 L vessel at 298 K.
\[[\text{H}_2\text{S}] = \_\_\_\_\_ \, \text{M} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e7279b6-864f-464c-a490-1592f397f56d%2Fde8005ae-d373-4d28-8f86-a5ea6f675e4d%2F7wv2qs_processed.png&w=3840&q=75)
Transcribed Image Text:**Equilibrium Constant Calculation**
The equilibrium constant, \( K_c \), for the following reaction is \(1.80 \times 10^{-4}\) at 298 K.
\[ \text{NH}_4\text{HS} (s) \rightleftharpoons \text{NH}_3 (g) + \text{H}_2\text{S} (g) \]
**Problem Statement:**
Calculate the equilibrium concentration of \(\text{H}_2\text{S}\) when 0.525 moles of \(\text{NH}_4\text{HS} (s)\) are introduced into a 1.00 L vessel at 298 K.
\[[\text{H}_2\text{S}] = \_\_\_\_\_ \, \text{M} \]
Expert Solution

Step 1: Defining equilibrium constant!
Answer:
For any reaction, value of equilibrium constant KC is equal to the ratio of molar concentration of products and reactants at equilibrium.
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