The initial rates method was used to study the reaction below. A + 3B - ProProfs Discuss
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The initial rates method was used to study the reaction below. A + 3B =2C [ (mol/L) [ (mol/L) -D[/Dt (mol/L×s) 0.210 0.150 3.41 x 10^3 0.210 0.300 1.36 x 10^2 0.420 0.300 2.73 x 10^-2



A. Rate = 0.515[ x [
B. Rate = 0.515[^2 x [
C. Rate = 0.721[^2 x [
D. Rate = 0.721[ x [^2
E. Rate = 0.721[^2 x [^2

This question is part of CHEM 2425
Asked by Ochiengj, Last updated: Feb 25, 2020

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1 Answer

John Smith

John Smith

Answered Feb 27, 2017

Rate = 0.721[A] x [B]^2

1.36 10-2/3.41 10-3= (0.210/0.210)m (0.300/0.150)n 4 = (2)n n = 2 2.73 10 -2/1.36 10-2 =(0.420/0.210)m (0.300/0.300)n 2 =(2)m m = 1 2.73 10 -2= k (0.420)m (0.300)n = k (0.420) (0.300)2 k = 0.721
 

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