Question

a certain first order reaction is 45% complete in 65s. what are the rate constant and half life for this process? I really want help with learning how to do this type of problem step by step im lost

Answer 1

So, we'd wanna use the Arrhenius equation: \(\large [A_t]=[A_0]*e^{-kt}\)

we'd have to make some assumptions, like that "45% complete" means that 45% of the reactant has been converted into something else.

Have you used this equation before?

Answer 2

so 55% remains therefore ln.55=-kt+ln1 im not sure what to do next alson not sure y the ln 1

Answer 3

yes, that's right.

you also specify the time to find k.

Then you use: \(t_{1/2}=\dfrac{ln(2)}{k}\)

Answer 4

not sure how to read ur notations

Answer 5

\(ln(0.55)=-k(65~s)+ln(1) \)

Answer 6

then half-life\(=t_{1/2}=\dfrac{ln(2)}{k}\)

Answer 7

do u divide -65 to get k

Answer 8

\(k=-\dfrac{ln(0.55)}{65~s}\)

Answer 9

if that's what you mean

Answer 10

im not sure what dfac means

Answer 11

are you not seeing the \(\LaTeX \)?

Answer 12

to get k, divide ln(0.55) by -65 sec

Answer 13

ok cool I got k now explain y I take ln(2)/k

Answer 14

explain why? it's a graphical result. It just works out because of the exponential model the equation uses.

Answer 15

well in my notes I have that t1/2=0.693/k

Answer 16

ln(2)=0.693

Answer 17

it's the same thing

Answer 18

ok cool totally didnt know what are the units

Answer 19

how long does it take for the reaction to go to ?percents

Answer 20

the units are based on what you found for k, it should be something like "/sec"

Answer 21

now that you know k, you can find how long it takes for the reaction to hit 50% or whatever

Answer 22

yes just need a start on the first one is 25%.

Answer 23

so set it up and i'll check it

Answer 24

i would im not sure i know how believe me i dont want you to just me the answer but i am lost or unaware of the process

Answer 25

is it the same way as above

Answer 26

yeah, it's the same as above except you're solving for t, not k