Chemistry, asked by mahtoganesh528, 4 months ago


equilibrium constant for
6.ox 10 2 il in a
reaction there
there are
of NH; present calculate the
concentration

Answers

Answered by Itzgoldenking
0

Answer:

Explanation:

For the equilibrium reaction 2HBr(g)⇌H  

2

​  

(g)+Br  

2

​  

(g) , the equilibrium constant is K=  

1.6×10  

5

 

1

​  

.

Let the decease in the equilibrium pressure of HBr be p bar.

The initial pressures of HBr, H  

2

​  

,Br  

2

​  

 are 10 bar, 0 bar and 0 bar respectively.

The equilibrium pressures are 10-p, p/2 and p/2 respectively.

The equilibrium constant expression is

K  

p

​  

=  

(P  

HBr

​  

)  

2

 

P  

H  

2

​  

 

​  

P  

Br  

2

​  

 

​  

 

​  

=  

(10−p)  

2

 

(p/2)×(p/2)

​  

=  

1.6×10  

5

 

1

​  

 

4(10−p)  

2

 

p  

2

 

​  

=  

1.6×10  

5

 

1

​  

 

400p=20−2p

p=0.0498 bar

The equilibium pressures are

P  

H  

2

​  

 

​  

=P  

Br  

2

​  

 

​  

=  

2

0.0498

​  

=0.0249 bar

P  

HBr

​  

=10−0.0498=10 bar

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