Explanation Given: B r 2 ( g ) + C l 2 ( g ) ⇌ 2 B r C l ( g ) Δ H for BrCl is 14 .6 kJ/mol Δ G for BrCl is -1 .0 kJ/mol Formula Used: ( i ) Δ G = Δ H − T Δ S ( i i ) Δ G r x n 0 = − R T ln K R i s t h e g a s c o n s t a n t , T i s t e m p e r a t u r e a t 298 K , k p i s t h e e q u i l i b r i u m f r o m t h e p a r t i a l p r e s s u r e . Calculation: The equilibrium constant can be calculated by, Δ H f 0 = 2 ( 14.6 k J ) − ( 30.91 k J ) − 1 ( 0 k J ) Δ H f 0 = − 1...

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Chapter 19, Problem 72E

Interpretation Introduction

(b)

To determine:

The above reaction to calculated can be under conditions.

$B r_{2}_{(g)} + C l_{2}_{(g)} ⇌ 2 B r C l_{(g)}$

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Chapter 19, Problem 71E

Chapter 19, Problem 73E

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(b) To determine: The above reaction to calculated can be under conditions. B r 2 ( g ) + C l 2 ( g ) ⇌ 2 B r C l ( g )

Solution Summary: The author calculates the equilibrium constant at 655 K for the following reactions: 2 N O 2 N 2 O 4

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(b)

To determine:

The above reaction to calculated can be under conditions.

$B r_{2}_{(g)} + C l_{2}_{(g)} ⇌ 2 B r C l_{(g)}$

Want to see the full answer?

Chapter 19 Solutions

(b) To determine: The above reaction to calculated can be under conditions. B r 2 ( g ) + C l 2 ( g ) ⇌ 2 B r C l ( g )

Solution Summary: The author calculates the equilibrium constant at 655 K for the following reactions: 2 N O 2 N 2 O 4

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(b) To determine: The above reaction to calculated can be under conditions. Br2(g)+Cl2(g)⇌2BrCl(g)

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Chapter 19 Solutions

(b)

To determine:

The above reaction to calculated can be under conditions.

$B r_{2}_{(g)} + C l_{2}_{(g)} ⇌ 2 B r C l_{(g)}$