122 exam 3 fa17-Compatibility Mode- Saved to this PC noert Dw Design Layout Refe

Question

122 exam 3 fa17-Compatibility Mode- Saved to this PC noert Dw Design Layout References Mailings Review View Help el me nhat you want to d Calibi (Body)11 A A Aa- AaBbCcD AaBbCr AaBbCc 1 Normal No Spe... Heading1 Heading2 Title Subtitie Paragragh s (15 points) INO1 Rate (M/s) 10 0.15 7.2 x 10 3.6 x 10 144 x 10 0.12 a) What is the rate law? b) What is the value of the rate constant? c) what is the rate when the initial concentrations are IN04 1-0.19M and INO, 1-003MP Type here to search 40 2 3 4 5 8 9

Explanation / Answer

let the rate law be (-r) = K[NH4+]a[ NO2-]b, where a and b are orders with respect to [NH4+] amd [NO2-] respectively and K is rate constant.

from 1st data point K[0.24]a [0.1]b= 7.2*10-4, (1)

from the second data point, K[0.12]a [0.15]b= 3.6*10-4 (2)

from the 3rd data point, K[0.12]a [0.30]b= 1.44*10-3 (3)

Eq.3/Eq.2 gives 2b= 4, b= 2,

Hence Eq.1 becomes K[0.24]a[ 0.1]2 = 7.2*10-4 (1)

Eq.3 becomes K[0.12]a [0.3]2= 1.44*10-3 (3)

Eq.1/Eq.3 gives (0.12/0.24)a [0.1/0.3]2 = 1/2

2a= 9/2

2a= 4.5

takinh ln

a ln2= ln(4.5)

a =2.16 ( order can be fraction as well )

the rate law is K[NH4+]2.16[NO2-]2

from Eq.1 K[0.24]2.16[0.1]2= 7.2*10-4

K= 1.5706/M3.16 sec

rate becomes 1.5706*[NH4+]2.16[ NO2-]2

at [NH4+]= 0.19M and [NO2-]=0.032M

rate = 1.5706*(0.19)2.16*(0.032)2=4.45*10-5M

(1/2)a = (1/2)*9

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