3H2(g)+N2rg)t)2NH3rg) +Heat -what is the Keqif the coscentratiors atequilibr'um are (Hz]=01 Mi [Nz]= 0.250 M and CNH3]=1.5om? PHow many gramsofnitrogen gas are needecd to produce 5OL of ammonia at STP?チら1
3H2(g)+N2rg)t)2NH3rg) +Heat -what is the Keqif the coscentratiors atequilibr'um are (Hz]=01 Mi [Nz]= 0.250 M and CNH3]=1.5om? PHow many gramsofnitrogen gas are needecd to produce 5OL of ammonia at STP?チら1
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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### Chemical Equilibrium and Reaction Stoichiometry
#### Problem 1:
Consider the reaction:
\[ \text{3H}_2\text{(g)} + \text{N}_2\text{(g)} \rightleftharpoons \text{2NH}_3\text{(g)} + \text{heat} \]
**Question:**
What is the equilibrium constant \( K_c \) if the concentrations at equilibrium are \([H_2] = 1.00 \, M\), \([N_2] = 0.250 \, M\) and \([NH_3] = 1.50 \, M\)?
---
#### Solution:
To find the equilibrium constant \( K_c \), we use the expression:
\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} \]
For the given reaction:
\[ K_c = \frac{[\text{NH}_3]^2}{[\text{H}_2]^3 [\text{N}_2]} \]
Substituting the given concentrations:
\[ K_c = \frac{(1.50 \, M)^2}{(1.00 \, M)^3 \cdot (0.250 \, M)} \]
Doing the math:
\[ K_c = \frac{2.25}{1 \cdot 0.25} \]
\[ K_c = \frac{2.25}{0.25} \]
\[ K_c = 9 \]
Thus, the equilibrium constant \( K_c \) is 9.
---
#### Problem 2:
**Question:**
How many grams of nitrogen gas (\( \text{N}_2 \)) are needed to produce 5.0 L of ammonia (\( \text{NH}_3 \)) at STP (Standard Temperature and Pressure)?
---
#### Solution:
At STP, 1 mole of any gas occupies 22.4 liters. First, we find the moles of ammonia produced:
\[ \text{moles of NH}_3 = \frac{5.0 \, L}{22.4 \, L/mol} \]
\[ \text{moles of NH}_3 = 0.223 \, mol \]
From the balanced equation, 1 mole of \( \text{N}_2 \) produces 2 moles of \( \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fac25bd78-e91e-4b13-9845-2104d21cc2ad%2Fb3c9d0b1-5eda-4842-82f0-b0bb0ace0675%2Fm7najel_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Chemical Equilibrium and Reaction Stoichiometry
#### Problem 1:
Consider the reaction:
\[ \text{3H}_2\text{(g)} + \text{N}_2\text{(g)} \rightleftharpoons \text{2NH}_3\text{(g)} + \text{heat} \]
**Question:**
What is the equilibrium constant \( K_c \) if the concentrations at equilibrium are \([H_2] = 1.00 \, M\), \([N_2] = 0.250 \, M\) and \([NH_3] = 1.50 \, M\)?
---
#### Solution:
To find the equilibrium constant \( K_c \), we use the expression:
\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} \]
For the given reaction:
\[ K_c = \frac{[\text{NH}_3]^2}{[\text{H}_2]^3 [\text{N}_2]} \]
Substituting the given concentrations:
\[ K_c = \frac{(1.50 \, M)^2}{(1.00 \, M)^3 \cdot (0.250 \, M)} \]
Doing the math:
\[ K_c = \frac{2.25}{1 \cdot 0.25} \]
\[ K_c = \frac{2.25}{0.25} \]
\[ K_c = 9 \]
Thus, the equilibrium constant \( K_c \) is 9.
---
#### Problem 2:
**Question:**
How many grams of nitrogen gas (\( \text{N}_2 \)) are needed to produce 5.0 L of ammonia (\( \text{NH}_3 \)) at STP (Standard Temperature and Pressure)?
---
#### Solution:
At STP, 1 mole of any gas occupies 22.4 liters. First, we find the moles of ammonia produced:
\[ \text{moles of NH}_3 = \frac{5.0 \, L}{22.4 \, L/mol} \]
\[ \text{moles of NH}_3 = 0.223 \, mol \]
From the balanced equation, 1 mole of \( \text{N}_2 \) produces 2 moles of \( \
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