Chemistry
posted by Elizabeth .
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)

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. 
2.33

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