2HI(g)--> H2(g) + I2(g)
The rate law for the decomposition of HI is r=k[HI]^2. The rate of the rxn was found experimentally to b 2.5 x 10^4 mol/L x s where the HI concentration was 0.0558 M.
Are these values ready to plug in to the equation, or because there is 2HI, do I have to divide the moles by 2?
I think the rate quoted is good as is; if any division/multiplication/etc are needed, they should have been done when quoting the rate.
In this reaction, the rate law is given as r = k[HI]^2, where [HI] represents the concentration of HI. The rate law tells you how the rate of the reaction depends on the concentration of the reactant.
In order to determine the value of the rate constant k, you need to use experimental data. In this case, you have been given the rate of the reaction (2.5 x 10^4 mol/L·s) and the initial concentration of HI (0.0558 M).
To find the value of k, you need to plug the given values into the rate law equation and solve for k.
2.5 x 10^4 mol/L·s = k * (0.0558 M)^2
Simplifying the equation, you get:
k = (2.5 x 10^4 mol/L·s) / (0.0558 M)^2
Now, let's address your question about the 2HI in the rate law equation. In this case, the stoichiometric coefficient of HI is 1, not 2, because the rate law is determined experimentally and may not necessarily match the balanced equation. The stoichiometric coefficient of 1 means that the concentration of HI should be raised to the power of 1 in the rate law equation. Therefore, you do not need to divide the concentration of HI by 2 in this case.
Once you have calculated the value of k, you can use it to determine the rate of the reaction for different concentrations of HI by plugging in the corresponding values into the rate law equation.