The following initial rate data are for the reduction of nitric oxide with hydro

Question

The following initial rate data are for the reduction of nitric oxide with hydrogen: 2 NO +2 H2N2+2 H20 Initial Ratc, Ms 5.16×10-2 0.206 0.103 0.413 Experiment NOlo, M 0.269 0.538 0.269 0.538 0.469 0.469 0.938 0.938 4 Complete the rate law for this reaction in the box below. Use the form kA mn , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter l for m or Rate = From these data, the rate constant is Submit Answer Retry Entire Group 6 more group attempts remaining

Explanation / Answer

1)
see experiment 1 and 2:
[NO] doubles
[H2] is constant
rate becomes 4 times
so, order of NO is 2

see experiment 1 and 3:
[NO] is constant
[H2] doubles
rate doubles
so, order of H2 is 1

Rate law is:
rate = k*[NO]^2*[H2]

Put values from 1st row of table in rate law
rate = k*[NO]^2*[H2]
5.16*10^-2 = k*0.269^2*0.469^1
k = 1.5205 M-2.s-1
Answer:
rate = k*[NO]^2*[H2]
1.52 M-2.s-1

2)
see experiment 1 and 3:
[C5H10] doubles
[O3] is constant
rate doubles
so, order of C5H10 is 1

see experiment 1 and 2:
[C5H10] is constant
[O3] doubles
rate doubles
so, order of O3 is 1

Rate law is:
rate = k*[C5H10]*[O3]

Put values from 1st row of table in rate law
rate = k*[C5H10]*[O3]
4.78*10^2 = k*0.178*0.031
k = 86625.589 M-1.s-1
Answer:
rate = k*[C5H10]*[O3]
86626 M-1.s-1

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