A certain first-order reaction has a rate constant of 2.80×10−2s−1 at 15∘C. What is the value of k at 62∘C if Ea = 112kJ/mol ?

Question

aaronq · Answer

use the two-point arrhenius equation:  $ln\dfrac{k_1}{k_2}=\dfrac{E_a}{R}*(\dfrac{1}{T_2}-\dfrac{1}{T_1})$

anonymous · Answer

I have tried that and it doesnt get the right answer :(

aaronq · Answer

you used Kelvin?

anonymous · Answer

yes and i changed kj to J

aaronq · Answer

let me see what i get

aaronq · Answer

are you getting 19.82?

anonymous · Answer

no -6.5

anonymous · Answer

do you have to do something with the k1 too?

aaronq · Answer

yeah, you have to fill all the parameters except what you're looking for

anonymous · Answer

cause if you do then i got .00004

aaronq · Answer

hm i got 19.82

anonymous · Answer

so 112000/8.314 * ((1/335)-(1/288)) = -6.56

aaronq · Answer

yeah, then    $ln(\dfrac{2.80×10^{−2}}{k_2})=-6.56$

anonymous · Answer

oh i see you take the reciprocal ok i got your answer now

aaronq · Answer

sweet, that should be correct

anonymous · Answer

but the answer is supossed to be 2.3

aaronq · Answer

really? thats odd. is this a reliable source?

anonymous · Answer

mastering chemistry...online text software

anonymous · Answer

oh no we were right!! I just answered the wrong question lol. Thanks a lot it is right!!

aaronq · Answer

ohh ok, cool cool

aaronq · Answer

no prob