Question

For the reaction: 2NO(g) + O_2(g) rightarrow 2NO_2(g) Determine the rate law: rate = k[NO]^m [O_2]^n Determine also the rate constant: k = (rate)/[NO]^m [O_2]^n

Answer 1

Rate law can be determined by comparing the given data.

Rate = k[NO]^m[O₂]ⁿ

taking log of the above equation,

Log rate = log k + m log [NO] + n log[O₂]

Substituting the values of each from the given data and forming respective equations

Equation a: log 0.048 = log k + m log 0.30 + n log 0.30

Equation b: log 0.192 = log k + m log 0.60 + n log 0.30

Equation c: log 0.768 = log k + m log 0.60 + n log 1.20

Subtracting equation a from equation b, we get

Log (0.192/0.048) = m log (0.60/0.30)

Log 4 = m log 2

2 log 2 = m log 2

m = 2

similarly, subtracting equation b from equation c, we get

Log (0.768/0.192) = n log (1.20/0.30)

Log 4 = n log 4

n = 1

now substituting value of m and n in equation a,

Log 0.048 = log k + 2 log 0.30 + log 0.30

Log k = log (0.048/(0.30×0.30×0.30))

k = 1.77

combining all the obtained values,

Rate law : rate = 1.77[NO]²[O₂]¹