Homework Help: Chemistry

Posted by Amphee on Tuesday, April 13, 2010 at 8:37pm.

The value of the equilibrium constant (Kp) as represented by the first chemical equation is 1.02 x 102 at 727 °C. Calculate the value of the equilibrium constant (Kp) for the second equation at the same temperature. Express answer in scientific notation.

2F(g) = F2(g)

F2(g) = 2F(g)

I'm not getting the math here, like how to mortify the first equation to equal it to the second one. [F]/[F]^2 = [F]^2/[F] ?

The answer is 9.8 x 10^-3

Chemistry - Amphee, Tuesday, April 13, 2010 at 8:47pm
Wait never mind I figured it out.
1/F = F so 1/102 right?
Chemistry - DrBob222, Tuesday, April 13, 2010 at 11:17pm
right.
1/k = 1/1.02 x 10^2 = 9.8 x 10^-3

Answer this Question

Related Questions

college chemistry - The value of the equilibrium constant (Kp) as represented by...
Chem - I am having problems with these two questions. 1.The value of the ...
chemistry - The value of the equilibrium constant (Kc) as represented by the ...
college chemistry - The value of the equilibrium constant, Kc, at a certain ...
Chemistry - The value of the equilibrium constant, Kc, at a certain temperature...
college chemistry - At 625 K the value of Kc for the equilibrium represented by ...
Chemistry (Equilibrium) - Given the reaction below 2NOBr(g) <==> ...
Chemistry - The equilibrium constant Kc for the following reaction is equal to 0...
chemistry - The equilibrium constant Kp for the following equilibrium is 33 at ...
chemistry - How do you figure out the answer for the reverse reaction? At 350 K ...

For Further Reading

First Name:
School Subject:
Answer: