The reaction of nitric oxide with hydrogen at 1280 c is 2no(g) +2h2(g)n2(g)+2h2o(g) from the following data collected at this temperature, determine the rate law and calculate the rate constant. initial rate (m/s) 1.3 x 10 5.0 x 10 experiment 1 2 3 no 0050 0100 0100 .0020 .0020 0040 10.0 x 105

Rate = 1.23*[NO]^2*[H2]

Explanation:

Data from question is unintelligible, so I'm going to use the table attached (the procedure doesn't change).

From experiments 1 and 2, when [H2] doubles, rate doubles; then, the order of reaction for [H2] is 1.

From experiments 2 and 3, when [NO] doubles, rate quadruples; then, the order of reaction for [NO] is 2.

The formula for the rate law is:

Rate = k*[NO]^2*[H2]

solving for the rate constant and replacing with the initial rate we get:

k = initial rate/([NO]^2*[H2])

k = 1.23*10^-3/(0.1^2*0.1) = 2.46*10^-3/(0.1^2*0.2) = 4.92*10^-3/(0.2^2*0.1)  = 1.23

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