Calculating value of Kp for variations of a given reaction such as reverse reaction, half quantities

For the reaction below, Kp = 6.0  104 at a certain temperature.

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

What is the value of Kp for the following reactions at this temperature?

    (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)

If someone can show me how to do one I'm pretty sure I can follow the same steps to do the other two.

Notice that reaction (a) is the reverse of the given reaction and is also 1/2 of the given reaction. When you reverse a reaction, the new value of K becomes 1/K or K-1. Also, when you multiply an equation by some factor (such as 1/2), then you raise K to that factor:
n ( A + B --> C)                  K
=n A  + n B ----n C              Kn

thanks, that's like the only thing I do understand ia that but what forumula am I using? what do I plug in for those values? because usually I am given the values for H, Br, and Hbr.

You don't need any of that stuff. Use the value of Kp they give you. So for reaction a, you see that they flipped the reaction, so do 1/K , where K is the value they give you. Now raise that to the 1/2 power and bingo, that's the value of K for reaction a. Similar methods can be applied to reactions b and c

so for a I would do...

1/(6.0x104)-1 ?

I'm inputing my answer but it is not matching.

thanks a lot by the way. =)

I think the problem is in your math.

Kp = 6 x 104 for the original equation.  So for reaction B, which is the reverse of the original reaction, Kpb will equal 1/Kp  (or Kp -1 which is the same thing).  By calculating Kpb as being equal to 1/Kp-1 you essentially said 1/(1/Kp) which is not correct.

Kpb = 1/Kp =  1 / (6.0 x 104) = 1.6 x 10 -5
This should be the correct answer for B.

Since A is 1/2 of B, you should be able to find the value of Kpa by raising Kpb to the 1/2 power (which means finding the square root of Kpb).

thanks.

nvm.

another question. how do I start this?

In a study of the following reaction at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr.

3 Fe(s) + 4 H2O(g) reverse reaction arrow Fe3O4(s) + 4 H2(g)

Calculate the value of Kp for this reaction at 1200 K. Hint: Apply Dalton's law of partial pressures