Can someone help explain to me this question? 14. Using the data in the table be
ID: 541733 • Letter: C
Question
Can someone help explain to me this question?
14. Using the data in the table below, determine the rate law, for: N2+ H2 + F2 N2H2F2. (1) Rate = k[N2][H2] (2) Rate = k[NJ[H2][F2] (3) Rate = k[H2][F2] (4) Rate = k[N2H,FJ (5) Rate = k[N][F2] Exp IN2] Hl Fl Initial Rate 10.10 M 0.10 M 0.20 M 0.040 M/s 2 0.20 M 0.20 M 0.20 M 0.080 M/s 3 0.10M 0.20 M 0.20 M 0.040 M/s 40.10 M 0.10 M 040 M 0.080 M/s 0.30 M 0.20 M 0.30M ?? 15, what is the missing rate in the table in #142 (1) 0.080 M/s 4) 0.240 Ms (5) 0.360 Ms 2) 0.120 M/s (3) 0.180 M/sExplanation / Answer
In order to calculate the rate law expression for a A+B reaction, we need to apply Initial Rates Method.
Note that the generic formula goes as follows:
r = k [A]^a [B]^b
Note that if we got at least 3 sets of point, in which we have A and B constant, then we could use:
r1 / r2 = (k1 [A]1^a [B]1^b) / (k2 [A]2^a [B]2^b)
If we assume K1 and K2 are constant, then K1= K2 cancel each other
r1 / r2 = ([A]1^a [B]1^b) / ( [A]2^a [B]2^b)
Then, order according to [A] and [B]
r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b
If we get two points in which A1 = A2, then we could get B, and vise versa for A...
From the data shown in YOUR table
choose point
2 and 3, so N2 is left
(0.08/0.04) = (0.2/0.1)^n
n = ln((0.08/0.04) ) / ln((0.2/0.1)) = 1st order with respect to n2
now..
choose point 1 and 2
(0.04/0.08) = (0.1/0.2)^1 * (0.1/0.2)^h * (1)
(0.04/0.08) /(0.1/0.2) = 0.5^h
ln(1) / ln(0.5) = h
h = 0 with respec tto H2
now choose point 3 and 4
(0.04/0.08) = (0.1/0.1)^1 * (1) ( 0.2/0.4)^f
ln(0.5) / ln(0.5) = f
f = 1
then
Rate = k*[N2][F2]
get k
choose point 1
(0.04) = k*(0.1)(1)(0.2)
k = 0.04/0.2 / 0.1 = 2
now
Rate = k*[N2][F2]
Rate = 2*(0.3)(0.3) = 0.18 M/s
from the answers given, choose point 3
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