Question

In: Chemistry

What is the rate law for the reaction? What is the reaction order?

2 NO2 (g) F2(g)2 NO2F(g) Trials [NO2] (M) [F2] (M) Initial Rate (M/s) ??96) 10-3 1.9 × 10-2 3.8 X 10-2 0.0170.064 0.064? 024 028 10x10 00240.064 0.128 - 2-t 0.034

What is the rate law for the reaction?

What is the reaction order?

What is the value of the rate constant, k?

Expert Solution

order w.r.t NO2

(a1/a2)^x = r1/r2

(0.017/0.034)^x = ((9.6*10^-3)/(1.9*10^-2))

x = 1

order w.r.t F2

(a2/a3)^y = r2/r3

(0.064/0.128)^y = ((1.9*10^-2)/(3.8*10^-2))

y = 1

reaction order = x+y = 1+1 = 2

rate law,

rate = k[NO2][F2]

rate constant(k) = rate/[NO2][F2]

k = (9.6*10^-3)/(0.017*0.064)

= 8.82 M-1.S-1

queen_honey_blossom answered 3 years ago

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