The equilibrium constant for the reaction: 2NO(g) + Br2(g)<----> 2NOBr(g) is Kc = 1.3x10^-2 at...
The equilibrium constant for the reaction: 2NO(g) + Br2(g) <----> 2NOBr(g) is Kc = 1.3x10^-2 at 1,000 K
a.) At this temperature, does the equilibrium favor the product or reactants?
b.) Calculate Kc for 2NOBr <----> 2NO + Br2
c.) Calculate Kc for NOBr <----> NO + 1/2Br2
Homework Answers
The concept used here is based on equilibrium constant (
) which applies where everything in the equilibrium reaction is in the same phase. This equilibrium constant is in terms of concentration which is represented by 
.
Firstly, the new state of equilibrium can be determined using the given condition. Then use the expression of for the other reactions also.
The general reaction in equilibrium is shown below.
The expression for the equilibrium constant can be written as follows.
![K
[C]*[D
[AT [B]](http://img.homeworklib.com/questions/8a41f370-ff35-11eb-8318-a725b075a54d.png?x-oss-process=image/resize,w_560)
…… (1)
Here, is the equilibrium constant for the reaction.
If 
, the reaction is in equilibrium.
If 
, the reaction will favor the reactants.
If 
, the reaction will favor the products
(1)
The given equilibrium reaction is shown below.
Write the equilibrium constant for this reaction using the equation (1):
![K - [NOBr]
[NO] [Br]
K=](http://img.homeworklib.com/questions/8c5cb850-ff35-11eb-8bc6-d5d664b9c4e8.png?x-oss-process=image/resize,w_560)
Substitute 
as the value of 
in the above equation.
![[NOBr]
-= 1.3x102
[NO] [Br]
[NOBr] = 1.3x10 ?([NO] [Br])](http://img.homeworklib.com/questions/8d652a10-ff35-11eb-a16b-3de0fc273d64.png?x-oss-process=image/resize,w_560)
…… (2)
(2)
The given equilibrium reaction is shown below.
Write the equilibrium constant for this reaction using the equation (1):
![[NO] [Br]
[NOBr]](http://img.homeworklib.com/questions/8e208410-ff35-11eb-995b-550c68596e1b.png?x-oss-process=image/resize,w_560)
Substitute ![[NOBr] =1.3x10 ([NO] [Br,]](http://img.homeworklib.com/questions/8e7c3da0-ff35-11eb-973b-23e8101444e1.png?x-oss-process=image/resize,w_560)
as the value of 
from equation (2) in this.
![[NO[Br, ]
* *1.3x10([NOT [Br,1)
K = 77](http://img.homeworklib.com/questions/8f1cdf90-ff35-11eb-851e-f755ad3be1c3.png?x-oss-process=image/resize,w_560)
(3)
The equilibrium reaction is shown below.
Write the equilibrium constant for this reaction using the equation (1):
![K. - [NO][Br.]
[NOBr]](http://img.homeworklib.com/questions/8fcb0660-ff35-11eb-bcc3-3f26e49708ad.png?x-oss-process=image/resize,w_560)
By squaring on both sides, the equation will be as follows.
![(K) - [NO][Br,]
NOBr?](http://img.homeworklib.com/questions/902adf90-ff35-11eb-8701-9724af46a2a6.png?x-oss-process=image/resize,w_560)
Substitute ![[NOBr]
=1.3x10-?([NO]”[Bry)](http://img.homeworklib.com/questions/9090b550-ff35-11eb-b02c-2be7919f3516.png?x-oss-process=image/resize,w_560)
as the value of from equation (II) in this.
![[NO][Br]
(K) =-
1.3x10²([NO][Br])
K = 777
K = 8.8](http://img.homeworklib.com/questions/91333da0-ff35-11eb-86c1-9135ac602fd2.png?x-oss-process=image/resize,w_560)
The equilibrium favors 
and 
.
The value of 
for 
is 
.
The value of for 
is 
.
Add Answer to:
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