equilibrium, anything will help such as a website with a related problem?

Question

for the following reaction, Kp=3.5E4 at 1495k
given equation H2+Br2>2HBr

what is the value of Kp for the following reaction at 1495k
(a)HBr>1/2(H2)+1/2(Br2)
(b)2HBr (c)1/2(H2)+1/2Br2>HBr

Answer 1

First of all let's see the expression of Kp for the first equation

H2+Br2 <=> 2HBr

Kp=P(HBr)^2 / (P(H2)*P(Br2))= 3.5*10^4

a) HBr <=>1/2H2+ 1/2Br2

Kp1=[P(H2)^1/2]*[P(Br2)^1/2] / P(HBr)
But if you look carefully Kp1=1/squareroot(Kp) =
=1/(SQRT(3.5*10^4))= 5.35*10^-3

Remember that since squareroot of x= x^1/2
Kp1=[P(H2)^1/2]*[P(Br2)^1/2] / P(HBr)=

= [P(H2)*P(Br2)/(P(HBr)^2)] ^(1/2)=

=1 / [P(HBr)^2/(P(H2)*P(Br2))] ^(1/2) =

=(1/Kp)^(1/2) = 1/Kp^(1/2)=1/Squareroot(Kp)

I am sorry I couldn't write it more clearly but it is a limitation of the site and we can't write equations in a better way.

b)2HBr <=> H2+ Br2
It is reverse of the initial reaction so

Kp2=(P(H2)*P(Br2)) / P(HBr)^2= 1/Kp= 1/(3.5*10^4)= 2.86*10^-5

c)1/2H2+1/2Br2 <=>HBr

is the reverse of a do Kp3=1/Kp1= 1/(5.35*10^-3)=1.87*10^2

Answer 2

Equillibrium problems can be understood here:

http://www.hyper-ad.com/tutoring/chemistry/chem_eqm1.htm
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/bases.php