Consider the following reaction where K. = 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g)
Consider the following reaction where K. = 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g)
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![### Reaction Equilibrium Analysis
Consider the following reaction where \( K_c = 1.80 \times 10^{-2} \) at 698 K.
\[ 2\text{HI}(g) \leftrightharpoons \text{H}_2(g) + \text{I}_2(g) \]
A reaction mixture was found to contain 0.279 moles of \(\text{HI}(g)\), \(4.64 \times 10^{-2}\) moles of \(\text{H}_2(g)\), and \(3.79 \times 10^{-2}\) moles of \(\text{I}_2(g)\), in a 1.00 liter container.
**Questions:**
1. Is the reaction at equilibrium?
2. If not, what direction must it run in order to reach equilibrium?
The reaction quotient, \( Q_c \), equals \(\boxed{\phantom{t}}\).
**The reaction**
A. must run in the forward direction to reach equilibrium.
B. must run in the reverse direction to reach equilibrium.
C. is at equilibrium.
To determine whether the reaction is at equilibrium, we need to compare the reaction quotient, \( Q_c \), with the equilibrium constant, \( K_c \).
The reaction quotient, \( Q_c \), is calculated using the expression:
\[ Q_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]
Given the concentrations:
- \([\text{HI}] = 0.279 \text{ M}\)
- \([\text{H}_2] = 4.64 \times 10^{-2} \text{ M}\)
- \([\text{I}_2] = 3.79 \times 10^{-2} \text{ M}\)
Plugging in these values:
\[ Q_c = \frac{(4.64 \times 10^{-2})(3.79 \times 10^{-2})}{(0.279)^2} \]
Calculate \( Q_c \) to compare it with \( K_c \). If \( Q_c < K_c \), the reaction must run in the forward direction. If \( Q_c > K_c \), it must run in the reverse direction. If \( Q_c = K_c \), the reaction is at equilibrium.
After calculating,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc7db68c-7828-4a06-bc34-18cdd71bf49b%2F2c68d278-4751-4071-8240-78d118bd8a36%2Fwo6ahyc_processed.png&w=3840&q=75)
Transcribed Image Text:### Reaction Equilibrium Analysis
Consider the following reaction where \( K_c = 1.80 \times 10^{-2} \) at 698 K.
\[ 2\text{HI}(g) \leftrightharpoons \text{H}_2(g) + \text{I}_2(g) \]
A reaction mixture was found to contain 0.279 moles of \(\text{HI}(g)\), \(4.64 \times 10^{-2}\) moles of \(\text{H}_2(g)\), and \(3.79 \times 10^{-2}\) moles of \(\text{I}_2(g)\), in a 1.00 liter container.
**Questions:**
1. Is the reaction at equilibrium?
2. If not, what direction must it run in order to reach equilibrium?
The reaction quotient, \( Q_c \), equals \(\boxed{\phantom{t}}\).
**The reaction**
A. must run in the forward direction to reach equilibrium.
B. must run in the reverse direction to reach equilibrium.
C. is at equilibrium.
To determine whether the reaction is at equilibrium, we need to compare the reaction quotient, \( Q_c \), with the equilibrium constant, \( K_c \).
The reaction quotient, \( Q_c \), is calculated using the expression:
\[ Q_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]
Given the concentrations:
- \([\text{HI}] = 0.279 \text{ M}\)
- \([\text{H}_2] = 4.64 \times 10^{-2} \text{ M}\)
- \([\text{I}_2] = 3.79 \times 10^{-2} \text{ M}\)
Plugging in these values:
\[ Q_c = \frac{(4.64 \times 10^{-2})(3.79 \times 10^{-2})}{(0.279)^2} \]
Calculate \( Q_c \) to compare it with \( K_c \). If \( Q_c < K_c \), the reaction must run in the forward direction. If \( Q_c > K_c \), it must run in the reverse direction. If \( Q_c = K_c \), the reaction is at equilibrium.
After calculating,
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