Explanation The required reaction is given below. 2 B r C l ( g ) ⇌ B r 2 ( g ) + C l 2 ( g ) , K = 32 The expression for equilibrium constant for the above reaction can be written as shown below, K = ( P B r 2 ) ( P C l 2 ) ( P B r C l ) 2 Given that, the initial partial pressure of B r C l is 3.30 m b a r . An equilibrium table can be set up as given below. C o m p o s i t i o n ( m b a r ) B r C l B r 2 C l 2 I n i t i a l p r e s s u r e 3.30 0 0 C h a n g e i n p r e s s u r e − 2 x + x + x E q u i l i b r i u m p r e s s u r e 3.30 − 2 x x x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below. K = ( x ) ( x ) ( 3

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

7th Edition

ISBN: 9781319403959

Author: ATKINS

Publisher: MAC HIGHER

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Chapter 5, Problem 5I.4AST

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Interpretation:

The equilibrium partial pressure of BrCl in the reaction mixture has to be calculated. The reaction is given below.

2BrCl(g)⇌Br2(g)+Cl2(g), K=32

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Chapter 5, Problem 5I.3BST

Chapter 5, Problem 5I.4BST

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Answers to the remaining 6 questions will be hand-drawn on paper and submitted as a single file upload below: Review of this week's reaction: H₂NCN (cyanamide) + CH3NHCH2COOH (sarcosine) + NaCl, NH4OH, H₂O ---> H₂NC(=NH)N(CH3)CH2COOH (creatine) Q7. Draw by hand the reaction of creatine synthesis listed above using line structures without showing the Cs and some of the Hs, but include the lone pairs of electrons wherever they apply. (4 pts) Q8. Considering the Zwitterion form of an amino acid, draw the Zwitterion form of Creatine. (2 pts) Q9. Explain with drawing why the C-N bond shown in creatine structure below can or cannot rotate. (3 pts) NH2(C=NH)-N(CH)CH2COOH This bond Q10. Draw two tautomers of creatine using line structures. (Note: this question is valid because problem Q9 is valid). (4 pts) Q11. Mechanism. After seeing and understanding the mechanism of creatine synthesis, students should be ready to understand the first half of one of the Grignard reactions presented in a past…

Propose a synthesis pathway for the following transformations. b) c) d)

Chapter 5 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

Ch. 5 - Prob. 5A.1AST Ch. 5 - Prob. 5A.1BST Ch. 5 - Prob. 5A.2AST Ch. 5 - Prob. 5A.2BST Ch. 5 - Prob. 5A.3AST Ch. 5 - Prob. 5A.3BST Ch. 5 - Prob. 5A.1E Ch. 5 - Prob. 5A.2E Ch. 5 - Prob. 5A.3E Ch. 5 - Prob. 5A.4E

Ch. 5 - Prob. 5A.5E Ch. 5 - Prob. 5A.6E Ch. 5 - Prob. 5A.7E Ch. 5 - Prob. 5A.8E Ch. 5 - Prob. 5A.11E Ch. 5 - Prob. 5B.1AST Ch. 5 - Prob. 5B.1BST Ch. 5 - Prob. 5B.2AST Ch. 5 - Prob. 5B.2BST Ch. 5 - Prob. 5B.3AST Ch. 5 - Prob. 5B.3BST Ch. 5 - Prob. 5B.1E Ch. 5 - Prob. 5B.2E Ch. 5 - Prob. 5B.3E Ch. 5 - Prob. 5B.5E Ch. 5 - Prob. 5B.7E Ch. 5 - Prob. 5C.1AST Ch. 5 - Prob. 5C.1BST Ch. 5 - Prob. 5C.2AST Ch. 5 - Prob. 5C.2BST Ch. 5 - Prob. 5C.3AST Ch. 5 - Prob. 5C.3BST Ch. 5 - Prob. 5C.1E Ch. 5 - Prob. 5C.3E Ch. 5 - Prob. 5C.4E Ch. 5 - Prob. 5C.5E Ch. 5 - Prob. 5C.6E Ch. 5 - Prob. 5C.7E Ch. 5 - Prob. 5C.8E Ch. 5 - Prob. 5C.9E Ch. 5 - Prob. 5C.10E Ch. 5 - Prob. 5C.11E Ch. 5 - Prob. 5C.12E Ch. 5 - Prob. 5C.15E Ch. 5 - Prob. 5C.16E Ch. 5 - Prob. 5D.1AST Ch. 5 - Prob. 5D.1BST Ch. 5 - Prob. 5D.1E Ch. 5 - Prob. 5D.2E Ch. 5 - Prob. 5D.3E Ch. 5 - Prob. 5D.4E Ch. 5 - Prob. 5D.5E Ch. 5 - Prob. 5D.6E Ch. 5 - Prob. 5D.7E Ch. 5 - Prob. 5D.8E Ch. 5 - Prob. 5D.9E Ch. 5 - Prob. 5D.10E Ch. 5 - Prob. 5D.11E Ch. 5 - Prob. 5D.12E Ch. 5 - Prob. 5D.13E Ch. 5 - Prob. 5D.14E Ch. 5 - Prob. 5D.15E Ch. 5 - Prob. 5D.16E Ch. 5 - Prob. 5D.18E Ch. 5 - Prob. 5D.19E Ch. 5 - Prob. 5D.20E Ch. 5 - Prob. 5E.1AST Ch. 5 - Prob. 5E.1BST Ch. 5 - Prob. 5E.2AST Ch. 5 - Prob. 5E.2BST Ch. 5 - Prob. 5E.1E Ch. 5 - Prob. 5E.2E Ch. 5 - Prob. 5E.11E Ch. 5 - Prob. 5E.12E Ch. 5 - Prob. 5F.1AST Ch. 5 - Prob. 5F.1BST Ch. 5 - Prob. 5F.2AST Ch. 5 - Prob. 5F.2BST Ch. 5 - Prob. 5F.3AST Ch. 5 - Prob. 5F.3BST Ch. 5 - Prob. 5F.4AST Ch. 5 - Prob. 5F.4BST Ch. 5 - Prob. 5F.5AST Ch. 5 - Prob. 5F.5BST Ch. 5 - Prob. 5F.1E Ch. 5 - Prob. 5F.2E Ch. 5 - Prob. 5F.3E Ch. 5 - Prob. 5F.5E Ch. 5 - Prob. 5F.7E Ch. 5 - Prob. 5F.9E Ch. 5 - Prob. 5F.10E Ch. 5 - Prob. 5F.11E Ch. 5 - Prob. 5F.12E Ch. 5 - Prob. 5F.13E Ch. 5 - Prob. 5F.14E Ch. 5 - Prob. 5F.15E Ch. 5 - Prob. 5F.16E Ch. 5 - Prob. 5G.1AST Ch. 5 - Prob. 5G.1BST Ch. 5 - Prob. 5G.2AST Ch. 5 - Prob. 5G.2BST Ch. 5 - Prob. 5G.3AST Ch. 5 - Prob. 5G.3BST Ch. 5 - Prob. 5G.4AST Ch. 5 - Prob. 5G.4BST Ch. 5 - Prob. 5G.5AST Ch. 5 - Prob. 5G.5BST Ch. 5 - Prob. 5G.1E Ch. 5 - Prob. 5G.2E Ch. 5 - Prob. 5G.3E Ch. 5 - Prob. 5G.4E Ch. 5 - Prob. 5G.7E Ch. 5 - Prob. 5G.8E Ch. 5 - Prob. 5G.9E Ch. 5 - Prob. 5G.11E Ch. 5 - Prob. 5G.12E Ch. 5 - Prob. 5G.13E Ch. 5 - Prob. 5G.14E Ch. 5 - Prob. 5G.15E Ch. 5 - Prob. 5G.16E Ch. 5 - Prob. 5G.17E Ch. 5 - Prob. 5G.19E Ch. 5 - Prob. 5G.20E Ch. 5 - Prob. 5G.21E Ch. 5 - Prob. 5G.22E Ch. 5 - Prob. 5H.1AST Ch. 5 - Prob. 5H.1BST Ch. 5 - Prob. 5H.2AST Ch. 5 - Prob. 5H.2BST Ch. 5 - Prob. 5H.1E Ch. 5 - Prob. 5H.2E Ch. 5 - Prob. 5H.3E Ch. 5 - Prob. 5H.4E Ch. 5 - Prob. 5H.5E Ch. 5 - Prob. 5H.6E Ch. 5 - Prob. 5I.1AST Ch. 5 - Prob. 5I.1BST Ch. 5 - Prob. 5I.2AST Ch. 5 - Prob. 5I.2BST Ch. 5 - Prob. 5I.3AST Ch. 5 - Prob. 5I.3BST Ch. 5 - Prob. 5I.4AST Ch. 5 - Prob. 5I.4BST Ch. 5 - Prob. 5I.1E Ch. 5 - Prob. 5I.2E Ch. 5 - Prob. 5I.3E Ch. 5 - Prob. 5I.4E Ch. 5 - Prob. 5I.5E Ch. 5 - Prob. 5I.6E Ch. 5 - Prob. 5I.7E Ch. 5 - Prob. 5I.9E Ch. 5 - Prob. 5I.10E Ch. 5 - Prob. 5I.11E Ch. 5 - Prob. 5I.12E Ch. 5 - Prob. 5I.13E Ch. 5 - Prob. 5I.14E Ch. 5 - Prob. 5I.15E Ch. 5 - Prob. 5I.16E Ch. 5 - Prob. 5I.17E Ch. 5 - Prob. 5I.18E Ch. 5 - Prob. 5I.19E Ch. 5 - Prob. 5I.20E Ch. 5 - Prob. 5I.21E Ch. 5 - Prob. 5I.22E Ch. 5 - Prob. 5I.23E Ch. 5 - Prob. 5I.24E Ch. 5 - Prob. 5I.25E Ch. 5 - Prob. 5I.26E Ch. 5 - Prob. 5I.27E Ch. 5 - Prob. 5I.28E Ch. 5 - Prob. 5I.29E Ch. 5 - Prob. 5I.30E Ch. 5 - Prob. 5I.32E Ch. 5 - Prob. 5I.33E Ch. 5 - Prob. 5I.34E Ch. 5 - Prob. 5I.35E Ch. 5 - Prob. 5I.36E Ch. 5 - Prob. 5J.1AST Ch. 5 - Prob. 5J.1BST Ch. 5 - Prob. 5J.3AST Ch. 5 - Prob. 5J.3BST Ch. 5 - Prob. 5J.4AST Ch. 5 - Prob. 5J.4BST Ch. 5 - Prob. 5J.5AST Ch. 5 - Prob. 5J.5BST Ch. 5 - Prob. 5J.1E Ch. 5 - Prob. 5J.2E Ch. 5 - Prob. 5J.3E Ch. 5 - Prob. 5J.4E Ch. 5 - Prob. 5J.5E Ch. 5 - Prob. 5J.6E Ch. 5 - Prob. 5J.9E Ch. 5 - Prob. 5J.10E Ch. 5 - Prob. 5J.11E Ch. 5 - Prob. 5J.12E Ch. 5 - Prob. 5J.13E Ch. 5 - Prob. 5J.17E Ch. 5 - Prob. 5.1E Ch. 5 - Prob. 5.2E Ch. 5 - Prob. 5.3E Ch. 5 - Prob. 5.4E Ch. 5 - Prob. 5.5E Ch. 5 - Prob. 5.6E Ch. 5 - Prob. 5.7E Ch. 5 - Prob. 5.8E Ch. 5 - Prob. 5.9E Ch. 5 - Prob. 5.10E Ch. 5 - Prob. 5.11E Ch. 5 - Prob. 5.12E Ch. 5 - Prob. 5.13E Ch. 5 - Prob. 5.14E Ch. 5 - Prob. 5.15E Ch. 5 - Prob. 5.16E Ch. 5 - Prob. 5.17E Ch. 5 - Prob. 5.19E Ch. 5 - Prob. 5.23E Ch. 5 - Prob. 5.24E Ch. 5 - Prob. 5.25E Ch. 5 - Prob. 5.26E Ch. 5 - Prob. 5.27E Ch. 5 - Prob. 5.28E Ch. 5 - Prob. 5.29E Ch. 5 - Prob. 5.30E Ch. 5 - Prob. 5.31E Ch. 5 - Prob. 5.32E Ch. 5 - Prob. 5.33E Ch. 5 - Prob. 5.35E Ch. 5 - Prob. 5.37E Ch. 5 - Prob. 5.38E Ch. 5 - Prob. 5.41E Ch. 5 - Prob. 5.43E Ch. 5 - Prob. 5.44E Ch. 5 - Prob. 5.45E Ch. 5 - Prob. 5.46E Ch. 5 - Prob. 5.47E Ch. 5 - Prob. 5.49E Ch. 5 - Prob. 5.51E Ch. 5 - Prob. 5.53E Ch. 5 - Prob. 5.55E Ch. 5 - Prob. 5.57E Ch. 5 - Prob. 5.58E Ch. 5 - Prob. 5.61E Ch. 5 - Prob. 5.62E

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The equilibrium partial pressure of B r C l in the reaction mixture has to be calculated. The reaction is given below. 2 B r C l ( g ) ⇌ B r 2 ( g ) + C l 2 ( g ) , K = 32

Solution Summary: The author explains how the equilibrium partial pressure of BrCl in the reaction mixture has to be calculated.

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Interpretation:

The equilibrium partial pressure of BrCl in the reaction mixture has to be calculated. The reaction is given below.

2BrCl(g)⇌Br2(g)+Cl2(g), K=32

Want to see the full answer?

Chapter 5 Solutions

The equilibrium partial pressure of B r C l in the reaction mixture has to be calculated. The reaction is given below. 2 B r C l ( g ) ⇌ B r 2 ( g ) + C l 2 ( g ) , K = 32

Solution Summary: The author explains how the equilibrium partial pressure of BrCl in the reaction mixture has to be calculated.

Videos

Interpretation: The equilibrium partial pressure of BrCl in the reaction mixture has to be calculated. The reaction is given below. 2BrCl(g)⇌Br2(g)+Cl2(g), K=32

Want to see the full answer?

Chapter 5 Solutions

Interpretation:

The equilibrium partial pressure of BrCl in the reaction mixture has to be calculated. The reaction is given below.

2BrCl(g)⇌Br2(g)+Cl2(g), K=32