The half life for the first order reaction a rightarrow B is 10.0 min. What perc
ID: 498147 • Letter: T
Question
The half life for the first order reaction a rightarrow B is 10.0 min. What percentage of A will react in 1.00 hour? (a) 1.56% (b) 93.3% (c) 1.00% (d) 99.0% (e) 98.4% Given the information in the previous question, what will be the reaction rate in 0.500 hour if initial concentration [A]_0 was 1.00 mM? (a) 8.66 times 10^-3 M min^-1 (b) 3.13 times 10^-5 M min^-1 (c) 8.66 times 10^-6 M min^-1 (d) 1.25 times 10^-4 M min^-1 (e) 1.25 times 10^-5 M min^-1 Cyclopropane is converted to propene in a first-order process. The rate constant is 5.4 times 10^-2 hr6-1. If is the initial concentration of cyclopropane is 0.15 M what percent of cyclopropane will remain after 100. min? (a) 8.6% (b) 93% (c) 7.0% (d) 91% (e) 0.45% Butadiene, C_4 H_6 (used to make synthetic rubber and latex paints) reacts to C_8 H_12 with a rate law of rate = 0.014 L/(mol middot s) [C_4 H_6]^2. What will be the concentration of C_4 H_6 after 3.0 hours if the initial concentration is 0.025 M? (a) 0.0052 M (b) 0.024 M (c) 43 M (d) 190 M (e) 0.0000 MExplanation / Answer
Answer 5)
For a first order reaction an integrated Rate Law is given as,
[A] = [Ao] x e(-kt). ------------------- (1)
And Half life (t1/2) and Rate constant (k) are related as,
k = 0.693/(t1/2). put in eq.(1)
[A] = [Ao] x e(-0.693t/(t1/2)).
For give First Order Reaction,
t1/2 = 10 min, t = 1 hr = 60 min
Using these values in eq.(1)
[A] = [Ao] x e(-0.693x60/10).
[A] = [Ao] x e(-4.158).
[A] = [Ao] x 0.0156
[A] / [Ao] = 0.0156
On multiplying by 100 on both sides
([A] / [Ao]) x 100 = 0.0156 x 100
%[A] = 1.56
1.56 % of [A] will reemains after 1 hour.
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