Hydrogen gas and bromine vapor react to form gaseous hydrogen bromide at 1300 K. The value of the equilibrium constant is 1.6 x 105. What is the value of the equilbrium constant for the decomposition of gaseous hydrogen bromide to form hydrogen gas and bromine vapor at 1300 K?

Question

anonymous · Answer

I am not quite sure, but isn't it just the reverse, because the reactants and products switch. I just think its logical, because of the way we calculate the equilibrium constant

anonymous · Answer

1 / 1.6x10^5 ?

anonymous · Answer

what do you mean they switch??

kainui · Answer

Well the formula for equilibrium constants is just products divided by reactants. So if you have something like this: $$H_2 + Br_2\rightarrow 2HBr$$ then that means $$2HBr \rightarrow H_2 + Br_2$$ is just the reverse reaction. So that means our rate constant which was originally calculated like this: $$k_1=\frac{ [HBr] }{ [H_2][Br_2] }$$

is very similar to the rate constant we want to calculate,
$$k_2=\frac{ [H_2][Br_2] }{[HBr]  }$$

and we can see that it's just the other one, but upside down. Make sense? The rate constant is just the ratio products to reactants.

anonymous · Answer

ok so thats why its the 1/6*10^5??

kainui · Answer

Yeah, exactly.

anonymous · Answer

ok thanks a bunch both of you!!!!!!