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crosbykins
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URGENT: chemistry qs: write the equillibrim constant for the reverse reaction
2. The equilibrium constant for the following reaction is 2.0 10^ 4
2HBr(g) H2(g) Br2(g)
b) What is the equilibrium constant for the reverse reaction?
I don't think this is right but
2.0 x 10^4 = (Br2)(H2)/(HBr)^2
2.0 x 10^4 = (x)(x)/(2x)^2
2.0 x 10^4 = 1/4
so for the reverse reaction K = 4 * 2.0 x 10^4.
Homework Statement
2. The equilibrium constant for the following reaction is 2.0 10^ 4
2HBr(g) H2(g) Br2(g)
Homework Equations
b) What is the equilibrium constant for the reverse reaction?
The Attempt at a Solution
I don't think this is right but
2.0 x 10^4 = (Br2)(H2)/(HBr)^2
2.0 x 10^4 = (x)(x)/(2x)^2
2.0 x 10^4 = 1/4
so for the reverse reaction K = 4 * 2.0 x 10^4.