Question

For the following reaction, Kp = 3.5 104 at 1495 K.

H2(g) + Br2(g) 2 HBr(g)

What is the value of Kp for the following reactions at 1495 K?
(a) HBr(g) 1/2 H2(g) + 1/2 Br2(g)
(b) 2 HBr(g) H2(g) + Br2(g)
(c) 1/2 H2(g) + 1/2 Br2(g) HBr(g)

Answer 1

For 1/2 the reaction, new k is sqrt Kp.
For 2 x reaction, new k is K^2p
For the reverse and 1/2 new k is 1/sqrt Kp.

Answer 2

2.33

Answer 3

a) K'p=(Kp)^{-1/2}=5.3*10^-3

b)K''p=(Kp)^{-1}=2.77*10^-5

c)K'''p=(Kp)^{1/2}=189.7

- Reversing the direction of a reaction gives the inverse of the equilibrium constant.
- When multiplying by a constant, raise K to the power of that constant