1. Consider the following generic reaction: 2A + B + C -> 2D +3E From the follow
ID: 881074 • Letter: 1
Question
1.
Consider the following generic reaction:
2A + B + C -> 2D +3E
From the following initial rate determinations, write the rate law for the reaction, determine the overall order for the reaction and calculate the rate constant, k.
Initial rate [A] [B] [C]
1.27x10^-4 Ms^-1 0.0125M 0.0125M 0.0125M
2.56x10^-4 Ms^-1 0.0250M 0.0125M 0.0125M
1.27x10^-4 Ms^-1 0.0125M 0.0250M 0.0125M
5.06x10^-4 Ms^-1 0.0125M 0.0125M 0.0250M
2. Consider the following reaction and its rate law:
2N2O5 -> 4NO2 + O2; rate = k[N2O5]
At 25 degrees Celsius, the half-life for this reaction is 96.3 minutes; calculate the initial N2O5 concentration, [N2O5]o, if [N2O5] = 1.80x10^-2 M after 60.0 minutes when the reaction proceeds at 25 degrees Celsius.
Explanation / Answer
1 )
rate= k [A]x [B]y [C]z
from data
1.27x10^-4 =k [0.0125]x [0.0125]y [0.0125]z ------------->(1)
2.56x10^-4 =k [0.0250]x [0.0125]y [0.0125]z ------------------> (2)
1.27x10^-4 =k [0.0125]x [0.0250]y [0.0125]z -------------> (3)
5.06x10^-4 =k [0.0125]x [0.0125]y [0.0250]z ----------------> (4)
by solving 1 and 2
x = 1
by solving 2 and 3
y = 1/2
by solving 3 and 4
z = 2
rate law = k [A]1 [B]1/2 [C]2
overall order = 1+ 1/2 + 2 = 3.5
from any of above equations we can solve rate constant k value
1.27x10^-4 =k [0.0125]x [0.0125]y [0.0125]z ------------->(1)
1.27x10^-4 =k [0.0125]1 [0.0125]1/2 [0.0125]2
k = 581.6 mol^-7/2 lit^7/2 sec^1
rate constant = k = 581.6 mol^-7/2 lit^7/2 sec^1
2)
rate constant = k = 0.693 / half life
= 0.693 / 96.3
= 7.196 x 10^-3 min^-1
for first order
initila concentration = A
after time t concentration = At
k = (2.303 / t ) * log (A / At)
7.196 x 10^-3 = (2.303 / 60 ) log (A /1.8 x10^-2)
0.1875 = log A - log (1.8 x10^-2)
0.1875 - 1.745 = log A
-1.557 = logA
A = 10^-1.557
A = 0.0277
A = 2.77 x 10^-2 M
initial concentration of N2O2 = 2.77 x 10^-2 M
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