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|References For the reaction N2(8) 2 H,0(8) 2 NO(g) +2 H3 (g) at 1100° C, the following data have been obtained. NO 4Na1/At r
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Answer #1

1)

see experiment 1 and 2:

[NO] doubles

[H2] is constant

rate becomes 4 times

so, order of NO is 2

see experiment 2 and 3:

[NO] is constant

[H2] doubles

rate doubles

so, order of H2 is 1

overall order = 2 + 1 = 3

Rate law is:

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

2)

Put values from 1st row of table in rate law

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

2*10^-2 = k*0.00593^2*0.38^1

k = 1497 M-2.s-1

Answer: 1.50*10^3 M-2.s-1

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