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

Chemical Principles: The Quest for Insight

7th Edition

ISBN: 9781464183959

Author: Peter Atkins, Loretta Jones, Leroy Laverman

Publisher: W. H. Freeman

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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…

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Chemical Principles: The Quest for Insight

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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