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Kinetics Question (spent over 7 hours on this) 
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#1
Jan3109, 09:46 AM

P: 8

1. The problem statement, all variables and given/known data
The following reaction was monitored as a function of time: AB> A + B A plot of 1/[AB] versus time yields a straight line with slope 5.2×10−2 M \s. If the initial concentration of AB} is 0.210 M, and the reaction mixture initially contains no products, what are the concentrations of {A} and {B} after 80 s? 2. Relevant equations 1/A = kt + 1 / A initial 3. The attempt at a solution 1/A = (5.2x10^2)(80 s) + 1/0.21 M *When I did this equation it was wrong. The correct answer was that the concentration of both A and B=9.8×10−2,9.8×10−2. Please tell me what I am doing wrong. Thanks! 


#2
Jan3109, 10:17 AM

P: 288

[tex]k=\frac{rate}{[AB]}[/tex]
and [AB] at t=0 can be used because k is a constant ratio. The rate is the slope of the line. However, the problem appears to have an error: [tex]\frac{1/[AB]}{t}=5.2x10^{2}M/s[/tex] The units should be [tex]\frac{1}{Ms}[/tex] If 1/[AB] vs. time is linear then k is not linear. Is this a first order reaction? 


#3
Jan3109, 10:40 AM

P: 8

Thanks for your response. I think the rate is second order because given was: Rate=k,[AB]^2
I just plugged in those numbers for first order and I am still getting the answer wrong. This question is frustrating me to the max. Any help would be hugely appreciated. 


#4
Jan3109, 10:58 AM

P: 288

Kinetics Question (spent over 7 hours on this)
Since it's a second order reaction, your equation for 1/[AB] is correct. Use conservation of mass. Your answer gives the concentration of AB after 80 seconds so this means what was converted is
.21  [AB] = 9.8 x 10^{2} M = concentration of products. 


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