Find the rate law and the overall reaction order for the reaction of H2 + I2 --> 2 Hi?

Question

Experiment is at 450oC
Experiment Initial Rate (mol/L.sec) [H2] [I2]
1) 1.9*10E-23 0.0113 0.001
2) 1.1*10E-22 0.0220 0.0033
3) 9.3*10E-23 0.0550 0.0022
4) 1.9*10E-22 0.0220 0.0056

I noted that H2 and I2 seem to be inverse, what else do you see?

Answer 1

Use number 2 and 4 to determine rate wrt I2, as H2 concentration is held constant.

1.9/1.1 = (0.0056/0.0033)^n
1.72 = 1.61^n

By this, essentially n = 1, so reaction is first order with respect to I2.

Knowing this, you can use any of the two to determine H2 concentration, using the ratio of one reaction to another. Using 1 and 2:

1.1*10^-22 = k[0.0220]^x[0.0033]
------------------------------------------
1.9*10^-23 = k[0.0113]^x[0.001]

5.79 = (1.95)^x*3
1.93 = (1.95)^x

Essentially, x is also 1, so reaction is first order wrt H2.

Rate law expression is then:

rate = k[H2][I2]

and the overall reaction order is second order.

Answer 2

the fee of any chemical reaction is the replace interior the concentration of a reactant (or product) in line with unit time, and that's what -d[I2]/dt refers to, i.e. the shrink (consequently the minus sign) interior the concentration of iodine with comprehend to time. you should the two properly write -d[H2]/dt (or +d[hi]/dt which may well be numerically two times as great because of the fact the different 2) all describe the fee of this reaction.