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 \( \
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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