A
reaction
A>B
observed. You do not know
the Stoichiometry of the reautent or
Prodent
but observe and when
you
increase the initial concentration
of
A front o.4M to
0.8 M, the half-life decreased from
10 minutes to 5 minutes - Determine the rate
to
concentration
Answers
Answer:lculate the fraction of the starting quantity of A that will be used up after 60 s. Given the reaction below which is found to be the first order in A and t1/2=40s
A→B+C(9.E.4)
S9.4
with \(t_{1/2} = 40 \; s\]
k=\frac{\ln 2}{t_{1/2}}=\frac{0.693}{40}=0.0173s^{-1}
The remain of fraction A after 60 s:
[A]=[A]_{o}e^{-kt}
\frac{[A]}{[A]_{o}}= e^{-(0.0173s^{-1})(60s)}= 0.354
The fraction will be used up after 60s:
1-0.354=0.646
Q9.5
Given the first order reaction is completed 90% in 30 mins at 298 K. Calculate the rate constant.
S9.5
90%= 0.9
[A]= [A]_{o}e^{-kt}
\frac{[A]}{[A]_{o}}=e^{-kt}
\ln \frac{[A]}{[A]_{o}}=-kt
\ln 0.9=-30k
k=-\frac{\ln 0.9}{30}=0.00351min^{-1}
Q9.6
Assume the half life of the first order decay of radioactive isotope takes about 1 year (365 days). How long will it take the radioactivity of that isotope to decay by 60%?
S9.6
t_{1/2}=\frac{\ln 2}{k}
k=\frac{\ln 2}{t_{1/2}}=\frac{0.693}{365}=0.00189day^{-1}
60% is lost
100%- 60%= 40% is remained
40%= 0.4
\frac{[A]}{[A]_{o}}=0.4
\frac{[A]}{[A]_{o}}=e^{-kt}
\ln \frac{[A]}{[A]_{o}}=-kt
t=\frac{-1}{k}\ln \frac{[A]}{[A]_{o}}=\frac{-1}{0.00189}\ln 0.4=484.81 days
Q9.7a
The decomposition of dinitrogen peroxide (N2O5) is a first-order reaction with a rate constant of 0.045 min-1 at 300 K.
2N2O5(g)→4NO2(g)+O2(9.E.5)
If there were initially 0.040 mol of N2O5, calculate the moles of N2O5 remaining after 5 minutes.
S9.7a
The integrated rate equation of a first-order reaction is:
[A]=[A]0e−kt(9.E.6)
Substituting concentration for moles of reactant and plugging in the known values:
nA=0.0400e−0.045∗5(9.E.7)
nA=0.032
Explanation:
hope it helps you