Part A The rate constant for a certain reaction is k 7.80x10-3 s-1. If the initi

Question

Part A The rate constant for a certain reaction is k 7.80x10-3 s-1. If the initial reactant concentration was 0.900 M, what will the concentration be after 20.0 minutes? Express your answer with the appropriate units. View Available Hint(s) Value Units Submit Part B er reaction has a constant rate of 3.30 10-4 M/s. If after 45.0 seconds the concentration has dropped to 4.00 102 M, what was the initial concentration? Express your answer with the appropriate units. View Available Hint(s) alue Units Submit

Explanation / Answer

part A) The unit of rate constant k is s^-1 ,so it must be a first order reaction.

k(unit)=(M/s)(1/M)^n .n=order of reaction

for n=1 k has unit s^-1

Using integrated rate law for first order reaction,

ln( [A]t/[A]o)=-kt

where t=time=20 min=20*60=1200 s

[A']t=final concentration of reactant after time t

[A]o=initial concentration of reactant=0.900M

k=rate constant=7.80*10^-3 s^-1

[A]t=[A]oexp(-kt)=(0.900M) exp((7.80*10^-3 s^-1)*(1200s))=7.749*10^-5 M

[A]t=7.749*10^-5 M (answer)

prt B) using integrated rate law for zero order rxn,

[A]t=-kt+[A]o

where t=time=45s

[A']t=final concentration of reactant after time t=4.00*10^-2 M

[A]o=initial concentration of reactant

k=rate constant=3.30*10^-4M/s

4.00*10^-2 M=-(3.30*10^-4M/s)(45s)+[A]o

or[A]o=4.00*10^-2 M- 0.01485M=0.02515

[A]o=0.0251M

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