18) If 2.00 moles of hydrogen and iodine are allowed to react in a10.0 L vessel according to the following reaction with K= 64.0 at a certain temperature, H2(g) + I2(g) <→2 HI(g), what will be the equilibrium concentration of HI? a) 0.16 M b) 0.32 M c) 0.48 M d) 0.64 M
18) If 2.00 moles of hydrogen and iodine are allowed to react in a10.0 L vessel according to the following reaction with K= 64.0 at a certain temperature, H2(g) + I2(g) <→2 HI(g), what will be the equilibrium concentration of HI? a) 0.16 M b) 0.32 M c) 0.48 M d) 0.64 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...
Related questions
Question
I need help please
![**Equilibrium Concentration Calculation Example**
**Problem Statement:**
If 2.00 moles of hydrogen (\(H_2\)) and iodine (\(I_2\)) are allowed to react in a 10.0 L vessel according to the following reaction with \( K_c = 64.0 \) at a certain temperature:
\[ H_2(g) + I_2(g) \leftrightarrow 2 HI(g) \]
What will be the equilibrium concentration of \(HI\)?
**Options:**
a) 0.16 M
b) 0.32 M
c) 0.48 M
d) 0.64 M
---
**Solution Approach:**
1. **Initial Concentrations:**
- Initial moles of \(H_2\) = 2.00 moles in 10.0 L -> Initial concentration of \(H_2\) = \(\frac{2.00}{10.0}\) = 0.20 M.
- Initial moles of \(I_2\) = 2.00 moles in 10.0 L -> Initial concentration of \(I_2\) = \(\frac{2.00}{10.0}\) = 0.20 M.
- Initial concentration of \(HI\) = 0 M (since none is initially present).
2. **Change in Concentrations:**
- Let \( x \) be the concentration of \( H_2 \) and \( I_2 \) that reacts.
- The reaction produces \( 2x \) moles of \( HI \).
- Thus, the change in concentration is:
\[ [H_2] = [I_2] = 0.20 - x \]
\[ [HI] = 2x \]
3. **Equilibrium Expression:**
- At equilibrium, substitute into the equilibrium expression for \( K_c \).
\[ K_c = \frac {[HI]^2}{[H_2][I_2]} = 64.0 \]
\[ \frac{(2x)^2}{(0.20 - x)(0.20 - x)} = 64.0 \]
4. **Solving for \( x \):**
- \( \frac {4x^2}{(0.20-x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd01c40b9-8944-4a44-8966-6ca7678f1a00%2F1ab211f0-247f-4edf-a52d-026fa04543a5%2Fpp67idf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Concentration Calculation Example**
**Problem Statement:**
If 2.00 moles of hydrogen (\(H_2\)) and iodine (\(I_2\)) are allowed to react in a 10.0 L vessel according to the following reaction with \( K_c = 64.0 \) at a certain temperature:
\[ H_2(g) + I_2(g) \leftrightarrow 2 HI(g) \]
What will be the equilibrium concentration of \(HI\)?
**Options:**
a) 0.16 M
b) 0.32 M
c) 0.48 M
d) 0.64 M
---
**Solution Approach:**
1. **Initial Concentrations:**
- Initial moles of \(H_2\) = 2.00 moles in 10.0 L -> Initial concentration of \(H_2\) = \(\frac{2.00}{10.0}\) = 0.20 M.
- Initial moles of \(I_2\) = 2.00 moles in 10.0 L -> Initial concentration of \(I_2\) = \(\frac{2.00}{10.0}\) = 0.20 M.
- Initial concentration of \(HI\) = 0 M (since none is initially present).
2. **Change in Concentrations:**
- Let \( x \) be the concentration of \( H_2 \) and \( I_2 \) that reacts.
- The reaction produces \( 2x \) moles of \( HI \).
- Thus, the change in concentration is:
\[ [H_2] = [I_2] = 0.20 - x \]
\[ [HI] = 2x \]
3. **Equilibrium Expression:**
- At equilibrium, substitute into the equilibrium expression for \( K_c \).
\[ K_c = \frac {[HI]^2}{[H_2][I_2]} = 64.0 \]
\[ \frac{(2x)^2}{(0.20 - x)(0.20 - x)} = 64.0 \]
4. **Solving for \( x \):**
- \( \frac {4x^2}{(0.20-x
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY