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

For the reaction below, Kp = 6.0 10^{4} 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.

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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 K

^{n}kyle1990

Sun, 2009-02-15 01:28

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.

Ohso

Sun, 2009-02-15 01:38

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

kyle1990

Sun, 2009-02-15 01:43

so for a I would do...

1/(6.0x10

^{4})^{-1}?I'm inputing my answer but it is not matching.

thanks a lot by the way. =)

Ohso

Sun, 2009-02-15 02:44

I think the problem is in your math.

Kp = 6 x 10

^{4}for the original equation. So for reaction B, which is the reverse of the original reaction, Kp_{b}will equal 1/Kp (or Kp -1 which is the same thing). By calculating Kp_{b}as being equal to 1/Kp-1 you essentially said 1/(1/Kp) which is not correct.Kp

_{b}= 1/Kp = 1 / (6.0 x 10^{4}) = 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 Kp

_{a}by raising Kp_{b}to the 1/2 power (which means finding the square root of Kpb).spock

Sun, 2009-02-15 08:19

thanks.

nvm.

Ohso

Sun, 2009-02-15 13:52

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

Ohso

Sun, 2009-02-15 16:06