The equilibrium constant at 1300 K for the reaction Ha(g)+ Br_2(g) 2HBr(g) is 1.6 times 10 The value of K for the reverse reaction is a. -1.6 times 10^5. b. 1.6 times 10^-5. c. 1.6 times 10^5 d. 6.3 times 10^-5 e. 6.3 times 10^-6.

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

Explanation:

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

6.4 cm (max) - 25 cm (min)