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Libby N

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Q: What is the rate constant of a reaction if rate = 1.5(mol/L)/s, [A] is 1M, [B] is 3 M, m=2, and n=1?
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Can rate of reaction be negative?

As far as I know, No. A negative order means a higher concentration of the reactant having a lower reaction rate. For example: concentration of A reaction rate ---- 4 M .1 M/s 1 M .4 M/s ---- rate1/rate2=k(A1)n/k(A2)n k canceled (.1M/s)/(.4M/s)=(4M/1M)n 1/4=4n n=-1 but the reaction rate is usually proportional to the concentration of the reactant, which means reactants with a higher concentration have a greater reaction rate, in a few case, increase the concentration of reactant have little effect for the reaction rate. So, the order of the reaction is usually positive in a few case, it's close to 0. For more information about the relationship between reaction rate and the concentration of recants check out: http:/en.wikipedia.org/wiki/Reaction_rate http:/www.chemguide.co.uk/physical/basicrates/concentration.html


For the reaction A to Products successive half lives are observed to be 10 min and 40 min A was 0.10 M at the beginning of the reaction What is the integrated rate law and what is the rate constant?

Half-life in this context is defined as the time needed to observe the halving of the original reactant concentration. As you can see, in this case that time doubles each time, so you can say that this particular reaction as a second order rate. The equation for this half-life is: t(1/2) = 1/([A]0 * k) and the integrated rate eq. is: 1/[A] = 1/[A]0 + kt Using the first half life we can calculate k: k = 1/( t(1/2) * [A]0) = 1/(10 min * 0.50 M) = 0.2 min^-1M^-1 Using this value we can calculate the answers for a) and b) a) 1/[A] = 1 / [A]0 + kt = 1 / 0.50 M + .2 min^-1M^-1 * 80 min = 18 M^-1 [A] = 0.056 M b) 1/[A] = 1 / [A]0 + kt = 1 / 0.50 M + .2 min^-1M^-1 * 10 min = 4 M^-1 [A] = 0.25 M


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