help work these chem questions (kinetics is a whore)

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C2H6 <-> 2CH3

at a temperature where the half-life is 38 minutes.
What is the first order rate contstant at this temperature?

.0182 min (got this)


How long would it take for the concentraton of C2H6 to drop from 4.5 mM to 0.6 mM?
t = ? min


If the initial concentration is 4.5 mM, what would the concentration be after 35.0 minutes?
[C2H6] = mM


How long would it take to convert 95% of C2H6 at this temperature?
t = min

 

[bullett]

What the fuck is C2H6?

 

[p0ink]

and why the hell is this wrong? im fucking doing it right!

2NOBr 2NO + Br2
The following concentration vs. time data was collected at some temperature:

t (s) [NOBr] (M)
0 .0615
5 .0480
15 .0334
25 .0256
40 .0189

k=slope= (delta(1/[A]))/delta t

so why isnt this right?

((1/.0189)-(1/.0615))/(40-0)=.9162

 

[toga22]

What the fuck is C2H6?

Ethane

 

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hheeeelllllppppppp

 

[aalox]

You just have to use the formula for exponential decay throughout this problem.
C = C(0) * exp(-Rt); where C is the momentary concentration and C(0) is the "initial" concentration. R is the rate of decay, which we can figure out from the half-life.
0.5 = 1 * exp(-R * 38m)
ln(0.5) = -R * 38m

* R = 0.0182 per minute

Now plug in the numbers for the second question.
0.6 mM = 4.5 nM * exp(-0.0182 * t)
(0.6 / 4.5) = exp(-0.0182 * t)
ln (0.6 / 4.5) / (-0.0182) = t

* t = 110.5 minutes

C = 4.5 mM * exp(-0.0182 * 35)

* C = 2.377 mM

The last question is similar to the second question.

5% = 100% * exp(-0.0182 * t)
0.05 = exp(-0.0182 * t)
ln (0.05) / (-0.0182) = t

* t = 164.2 minutes

 

[aalox]

and why the hell is this wrong? im fucking doing it right!

2NOBr 2NO + Br2
The following concentration vs. time data was collected at some temperature:

t (s) [NOBr] (M)
0 .0615
5 .0480
15 .0334
25 .0256
40 .0189

k=slope= (delta(1/[A]))/delta t

so why isnt this right?

((1/.0189)-(1/.0615))/(40-0)=.9162

You have to take the logarithm of the observed concentrations before you try to determine the slope. Plotting the logs will give you a reasonably straight line, while plotting the concentrations will give you an exponential curve.

 

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thanks a ton man. i hit you up with some green.