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jack ross
Questions (3)
i posted this question a few minutes ago i tried to solve it on my own can someone check if my work is okay?
The data in the
6 answers
581 views
The data in the table below were obtained for the reaction:
A + B → P 3 Experiments 1 (A) (M): 0.273 (B) (M): 0.763 Initial
3 answers
2,488 views
The data in the table below were obtained for the reaction:
A + B → P 3 Experiments 1 (A) (M): 0.273 (B) (M): 0.763 Initial
1 answer
887 views
Answers (7)
is this right? For first order kinetics: ln[A) = -kt + ln[Ao] ln[A) = - (6.29 x 10^-4 1/s)(1000. s) + ln (0.00100) ln[A) = -0.629 - 6.91 = -7.54 [A) = e^-7.54 = 5.33 x 10^-4
thanks for clearing it up, i understand it now.
but my book says x=2 ...?
oh, im sorry our teacher told us to use log in problems like these he didn't mention anything about exponents.
also, there is one more question that is asking me what the magnitude of the rate constant is how do i figure that out?
i did this: 2) 2.83 = k*[0.273]x[1.526]y ---------------------------------------- 1) 2.83 = k*[0.273]x[0.763]y 2.83=2y log: 0.405 = y ------- 0.301 1.5=y 2=y is that right too?
i divided: 1.526/0.763 and it gives me 2 2.83=2y
