# For the reaction #2"NO"(g) + "Br"_2(g) -> 2"NOBr"(g)#, the following data were obtained. What is the rate constant?

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#ul("Trial"" "["NO"]" "" "["Br"_2]" "color(white)("///")r(t) ("M/s"))#

#1color(white)("//////")"0.10 M"color(white)("///")"0.20 M"color(white)(///)24#

#2color(white)("//////")"0.25 M"color(white)("///")"0.20 M"color(white)(///)150#

#3color(white)("//////")"0.10 M"color(white)("///")"0.50 M"color(white)(///)60#

##### 1 Answer

The *general* rate law would be

#r(t) = k[A]^m[B]^n#

#= -1/a (Delta[A])/(Deltat) = -1/b (Delta[B])/(Deltat) = 1/c (Delta[C])/(Deltat) = 1/d (Delta[D])/(Deltat)# ,

for the reaction

#aA + bB -> cC + dD# ,

where:

#r(t)# is the initial rate of reaction.#k# is the rate constant for aparticularreaction at aparticulartemperature.#[X]# is the initial concentration of#X# .#m# and#n# are orders with respect to#A# and#B# , respectively. That is, they are the exponents that describe the contribution of#A# or#B# to the overall rate.

Therefore, the **rate law** for the reaction

#2NO(g) + Br_2(g) -> 2NOBr(g)#

is:

#r(t) = k[NO]^m[Br_2]^n#

An initial rate

#r_1(t) = k[NO]_1^m[Br_2]_1^n#

#r_2(t) = k[NO]_2^m[Br_2]_2^n#

#r_3(t) = k[NO]_3^m[Br_2]_3^n#

Since we do not know the rate constant, we can simply ignore it for the moment by dividing the rates. It is convenient to choose trials such that one of the concentrations remains constant across trials, eliminating a variable.

#(r_1(t))/(r_2(t)) = (cancel(k)[NO]_1^m[Br_2]_1^n)/(cancel(k)[NO]_2^m[Br_2]_2^n)#

#= (([NO]_1)/([NO]_2))^mcancel((([Br_2]_1)/([Br_2]_2))^n)^(1)#

We eliminated the bromine concentrations since

#ln[(r_1(t))/(r_2(t))] = ln(([NO]_1)/([NO]_2))^m#

#= mln(([NO]_1)/([NO]_2))#

#=> m = (ln[(r_1(t))/(r_2(t))])/(ln(([NO]_1)/([NO]_2)))#

However, your ratios are nice. From trials 1 and 2,

"What is

#0.4# raised to so that it equals#0.16# ?" That is, "what exponent makes it so that the ratio of the concentrations equals the ratio of the rates?"

Naturally, the answer is

#color(green)(m) = (ln0.16)/(ln0.4) = ln(0.4^2)/(ln(0.4)) = (2cancel(ln0.4))/(cancel(ln0.4)) = color(green)(2)#

For the other order

Or the brute force way:

#color(green)(n) = (ln[(r_1(t))/(r_3(t))])/(ln(([Br_2]_1)/([Br_2]_3))) = (ln0.4)/(ln0.4) = color(green)(1)#

Therefore, the rate law becomes:

#color(green)(r(t) = k[NO]^2[Br_2])#

Finally, we have the info to plug and chug. Choose any trial you wish. You will get effectively the same

#color(blue)(k) = (r_1(t))/([NO]_1^2[Br_2]_1)#

#= ("24 M/s")/(("0.10 M")^2("0.20 M"))#

#= color(blue)(1.2 xx 10^4)#

What must the units be?