Determine the rate law according to the following data. [H2] (M) [NO] (M) Rate (M/s) Trial 1 0.35 0.30 2.835 x 10-3 Trial 2 0.35 0.60 1.134 x 10-2 Trial 3 0.70 0.60 2.268 x 10-2

rate = k [H2O] [NO]²

Explanation:

[H2] (M) [NO] (M) Rate (M/s)

Trial 1 0.35 0.30 2.835 x 10-3

Trial 2 0.35 0.60 1.134 x 10-2

Trial 3 0.70 0.60 2.268 x 10-2

Comparing Trial 1 and 2 , the rate of the reaction increases by a factor of 4 when the concentration of NO was doubled. This means that the reaction is second order with respect to NO

Comparing Trial 2 and Trial 3, the rate of reaction increased by a factor of 2 when the concentration of H2O was doubled. This means that the reaction is first order with respect to H2O

The rate law fr the reaction is given as;

rate = k [H2O] [NO]²

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.