\"For the reaction 2 N2O5 --> 4 NO2 + O2 the rate constant is 6.82 x 10^-3 s^-1
ID: 914228 • Letter: #
Question
"For the reaction 2 N2O5 --> 4 NO2 + O2 the rate constant is 6.82 x 10^-3 s^-1 at 70 degrees Celsius. The reaction is first order overall. If you start with 0.350 mol of dinitrogen pentoxide in a 2.0 L volume, how many miles will remain after 10 minutes? How long will it take for you to have 0.125 moles of reactant left? What is the half-life of dinitrogen pentoxide?" *Please show all work. * if it's out of 2 L would you have to change to concentration (in molarity) and divide 0.350 mols by 2 or does that not matter?"For the reaction 2 N2O5 --> 4 NO2 + O2 the rate constant is 6.82 x 10^-3 s^-1 at 70 degrees Celsius. The reaction is first order overall. If you start with 0.350 mol of dinitrogen pentoxide in a 2.0 L volume, how many miles will remain after 10 minutes? How long will it take for you to have 0.125 moles of reactant left? What is the half-life of dinitrogen pentoxide?" *Please show all work. * if it's out of 2 L would you have to change to concentration (in molarity) and divide 0.350 mols by 2 or does that not matter?
*Please show all work. * if it's out of 2 L would you have to change to concentration (in molarity) and divide 0.350 mols by 2 or does that not matter?
Explanation / Answer
Given that;
The reaction is first order overall
a)
The formula for a 1st order reaction is:
ln(Af/Ao) = -kt
Here Af = final concentration and A o initial concentration, k rate constant and t = time
rearranging to solve for the ration gives
(Af/Ao) = e^(-kt)
The time can be converted to seconds:
t = (10.0 min) x [(60 s)/(1 min)] = 600 s
Plugging in all of the values gives:
(Af/Ao) = e^[-(6.82x10^-3 s-1)(600 s)]
(Af/Ao) = e^[-4.092]
(Af/Ao) = 0. 017
Af = 0.017 Ao
Af = 0.017 (0.350 moles) =0.00585 moles
b)
The formula for a 1st order reaction is:
ln(Af/Ao) = -kt
rearranging to solve for the ration gives
(Af/Ao) = e^(-kt)
Plug the values into the rate equation:
ln(0.125/0.350 ) = -(6.82x10^-3 s-1)t
- 1.03 = -(6.82x10^-3 s-1)t
t = 151 s
c)
the half-life is given by:
t(1/2) = 0.693/k
t(1/2) = 0.693/(6.82x10^-3 s-1)
= 102 s
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