Given reactions are: 2NH3(9) 1. N2(g) + 3H2(g) 2. H2(g) + Cl2(g) 2HC 3. 2CO2(g) + O2(g) → 2CO2(g) Calculate the standard entropy values for the following reactions at 25°C.
Given reactions are: 2NH3(9) 1. N2(g) + 3H2(g) 2. H2(g) + Cl2(g) 2HC 3. 2CO2(g) + O2(g) → 2CO2(g) Calculate the standard entropy values for the following reactions at 25°C.
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
Chapter17: Spontaneity, Entropy, And Free Energy
Section: Chapter Questions
Problem 5RQ: What is the third law of thermodynamics? What are standard entropy values, S, and how are these S...
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Hello! I just want to ask for help whether the answers in the given pictures are correct. If it's not, please help me recheck and resolve it. Please refer to the given pictures below for the answers and questions. Please read the instructions and directions very carefully.
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![Step 1
We are given reaction :
4) 302 (g)
NH2CONH2 (aq)
+ H₂0 (1)
5) 2 NH3(g)
CO2 (g)
Step 2
4) Given reaction is :
302 (g)
203 (g)
As,
AS⁰ = [Sum of entropy of formation of products] - [Sum of entropy of formation of
reactants]
Then,
AS⁰ = [2 x AS°f of 03 (g) ] - [3 × AS°ƒ of O2 (g)]
= [2 x 238.93 J/(K.mol)] - [3 x 205.14 J/(K.mol)]
= -137.56 J/(K.mol)
AS° -137.56 J/(K.mol)
Step 3
5) Given reaction is :
2 NH3(g)
+ H₂0 (1)
+ CO2(g)
NH2CONH2 (aq)
As,
AS⁰ = [Sum of entropy of formation of products] - [Sum of entropy of formation of
reactants]
Then,
2
AS⁰ = [AS°f of NH₂CONH2 (aq) + AS°f of H₂O(1) ]-[2x AS°f of NH3(g) + AS°f of CO₂ (g) ]
= [104.60 J/(K.mol) + 69.91 J/(K.mol) ] - [2 x 192.45 J/(K.mol) + 213.74 J/(K.mol)]
= -424.13 J/(K.mol)
AS = -424.13 J/(K.mol)
+
203 (g)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92a0eb1b-c8e1-49c9-9609-54c6a81f749c%2F311bea86-efe5-4771-bac8-1ccddd82f0ba%2F6bezr6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Step 1
We are given reaction :
4) 302 (g)
NH2CONH2 (aq)
+ H₂0 (1)
5) 2 NH3(g)
CO2 (g)
Step 2
4) Given reaction is :
302 (g)
203 (g)
As,
AS⁰ = [Sum of entropy of formation of products] - [Sum of entropy of formation of
reactants]
Then,
AS⁰ = [2 x AS°f of 03 (g) ] - [3 × AS°ƒ of O2 (g)]
= [2 x 238.93 J/(K.mol)] - [3 x 205.14 J/(K.mol)]
= -137.56 J/(K.mol)
AS° -137.56 J/(K.mol)
Step 3
5) Given reaction is :
2 NH3(g)
+ H₂0 (1)
+ CO2(g)
NH2CONH2 (aq)
As,
AS⁰ = [Sum of entropy of formation of products] - [Sum of entropy of formation of
reactants]
Then,
2
AS⁰ = [AS°f of NH₂CONH2 (aq) + AS°f of H₂O(1) ]-[2x AS°f of NH3(g) + AS°f of CO₂ (g) ]
= [104.60 J/(K.mol) + 69.91 J/(K.mol) ] - [2 x 192.45 J/(K.mol) + 213.74 J/(K.mol)]
= -424.13 J/(K.mol)
AS = -424.13 J/(K.mol)
+
203 (g)
![Given reactions are:
1.
N2(g) + 3H2(g)
2NH3(g)
2. H2(g) + Cl2(g) → 2HCl
3.
2CO2(g) + O2(g) → 2CO2(g)
1
Calculate the standard entropy values for the following reactions at 25°C.
Step 2
Calculation of standard enthalpy of the reactions (AHº) :-
1. Given reaction is:
N₂ (g) + 3H₂ (g)--------->2NH3(g)
As,
AHO = [Sum of enthalpy of formation of products] - [Sum of enthalpy of formation of reactants]
Then,
AH° = [2 x AHºf of NH3 (g)] - [AH° of N₂ (g) + 3 x AH°f of H₂ (g)]
= [2 x (-45.9 KJ/mole)] - [0 KJ/mole + 3 x 0 KJ/mole]
= -91.8 KJ
Therefore, AH°= -91.8 KJ
Step 3
2. Given reaction is:
H₂ (g) + Cl₂ (g) -----------> 2HCl (g)
As,
AHO = [Sum of enthalpy of formation of products] - [Sum of enthalpy of formation of reactants]
Then,
AH° = [2 x AH° of HCL (g)] - [AH°f of H₂ (g) + AH°f of Cl₂ (g)]
Then,
AH = [2 x (-92.3 KJ/mole)] - [0 KJ/mole + 0 KJ/mole]
- 184.6 KJ/mole
Hence, AH = - 184.6 KJ/mole
Step 4
3. Given reaction is:
2CO (g) + O₂ (g) -----------> 2CO₂ (g)
As,
AHO = [Sum of enthalpy of formation of products] - [Sum of enthalpy of formation of reactants]
Then,
AH° = [2 x AHºf of CO₂ (g)] - [2 x AHºf of CO (g) + AH° of O₂ (g)]
Then,
AH° [2x()]-[2 x () + 0 KJ/mole]
= [2 x (-393.5 KJ/mole)] - [2 x (-110.5 KJ/mole)]
= -787 KJ/mole +221 KJ/mole
= -566 KJ/mole
Hence, AH° = -566 KJ/mole](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92a0eb1b-c8e1-49c9-9609-54c6a81f749c%2F311bea86-efe5-4771-bac8-1ccddd82f0ba%2Ff7puaat_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given reactions are:
1.
N2(g) + 3H2(g)
2NH3(g)
2. H2(g) + Cl2(g) → 2HCl
3.
2CO2(g) + O2(g) → 2CO2(g)
1
Calculate the standard entropy values for the following reactions at 25°C.
Step 2
Calculation of standard enthalpy of the reactions (AHº) :-
1. Given reaction is:
N₂ (g) + 3H₂ (g)--------->2NH3(g)
As,
AHO = [Sum of enthalpy of formation of products] - [Sum of enthalpy of formation of reactants]
Then,
AH° = [2 x AHºf of NH3 (g)] - [AH° of N₂ (g) + 3 x AH°f of H₂ (g)]
= [2 x (-45.9 KJ/mole)] - [0 KJ/mole + 3 x 0 KJ/mole]
= -91.8 KJ
Therefore, AH°= -91.8 KJ
Step 3
2. Given reaction is:
H₂ (g) + Cl₂ (g) -----------> 2HCl (g)
As,
AHO = [Sum of enthalpy of formation of products] - [Sum of enthalpy of formation of reactants]
Then,
AH° = [2 x AH° of HCL (g)] - [AH°f of H₂ (g) + AH°f of Cl₂ (g)]
Then,
AH = [2 x (-92.3 KJ/mole)] - [0 KJ/mole + 0 KJ/mole]
- 184.6 KJ/mole
Hence, AH = - 184.6 KJ/mole
Step 4
3. Given reaction is:
2CO (g) + O₂ (g) -----------> 2CO₂ (g)
As,
AHO = [Sum of enthalpy of formation of products] - [Sum of enthalpy of formation of reactants]
Then,
AH° = [2 x AHºf of CO₂ (g)] - [2 x AHºf of CO (g) + AH° of O₂ (g)]
Then,
AH° [2x()]-[2 x () + 0 KJ/mole]
= [2 x (-393.5 KJ/mole)] - [2 x (-110.5 KJ/mole)]
= -787 KJ/mole +221 KJ/mole
= -566 KJ/mole
Hence, AH° = -566 KJ/mole
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