Question

If the concentration of a reactant is 0.0560 M 25.5 seconds after a reaction starts and is 0.0156 M 60.5 seconds after the start of the reaction , how many seconds after the start of the reaction does it take for the reactant concentration to decrease to 0.00450 M? Assume the reaction is first order in this reactant.

Answer #1

The reactant concentration in a zero-order reaction was
6.00×10−2 M after 160 s and 1.50×10−2 M after 305 s . What is the
rate constant for this reaction? (I got this answer 3.10*10^-4 M/s)
What was the initial reactant concentration for the reaction
described in Part A?
The reactant concentration in a first-order reaction was 0.100
M after 40.0 s and 3.80×10−3M after
90.0 s . What is the rate constant for this reaction? (I got this
answer 6.54*10^-2 1/s)...

The reactant concentration in a first-order reaction was
7.40×10−2 M after 10.0 s and 9.70×10−3 M after 100 s . What is the
rate constant for this reaction?The reactant concentration in a
second-order reaction was 0.800 M after 180 s and 1.60×10−2 M after
890 s . What is the rate constant for this reaction?

C) The reactant concentration in a first-order reaction was
8.40×10−2 M after 30.0 s and 1.50×10−3 M after 100 s . What is the
rate constant for this reaction?
D)The reactant concentration in a second-order reaction was
0.320 M after 250 s and 7.40×10−2 M after 750 s . What is the rate
constant for this reaction? Express your answer with the
appropriate units. Include an asterisk to indicate a compound unit
with mulitplication, for example write a Newton-meter as...

The integrated rate law allows chemists to predict the reactant
concentration after a certain amount of time, or the time it would
take for a certain concentration to be reached. The integrated rate
law for a first-order reaction is: [A]=[A]0e−kt Now say we are
particularly interested in the time it would take for the
concentration to become one-half of its initial value. Then we
could substitute [A]02 for [A] and rearrange the equation to:
t1/2=0.693k This equation calculates the time...

The integrated rate law allows chemists to predict the reactant
concentration after a certain amount of time, or the time it would
take for a certain concentration to be reached. The integrated rate
law for a first-order reaction is: [A]=[A]0e−kt Now say we are
particularly interested in the time it would take for the
concentration to become one-half of its inital value. Then we could
substitute [A]02 for [A] and rearrange the equation to: t1/2=0.693k
This equation caculates the time...

a.) The rate constant for the reaction is 0.460
M–1·s–1 at 200 °C. A--> products. If the
initial concentration of A is 0.00680 M. what is the concentration
after 315 s?
b.)The rate constant for this zero-order reaction
is 0.0190 M·s–1 at 300 °C. A--> products. How
long (in seconds) would it take for the concentration of A to
decrease from 0.800 M to 0.240 M?
c.)The rate constant for this second-order
reaction is 0.520 M–1·s–1 at 300 °C. A-->...

a.) The rate constant for the reaction is 0.460
M–1·s–1 at 200 °C. A--> products. If the
initial concentration of A is 0.00680 M. what is the concentration
after 315 s?
b.)The rate constant for this zero-order reaction
is 0.0190 M·s–1 at 300 °C. A--> products. How
long (in seconds) would it take for the concentration of A to
decrease from 0.800 M to 0.240 M?
c.)The rate constant for this second-order
reaction is 0.520 M–1·s–1 at 300 °C. A-->...

The reactant concentration in a second-order reaction was 0.110
M after 235 s and 2.70×10−2 M after 725 s . What is the rate
constant for this reaction? Express your answer with the
appropriate units. Include an asterisk to indicate a compound unit
with multiplication, for example write a Newton-meter as N*m.

Suppose a reaction with a single reactant is first order in that
reactant. As a first-order reaction, the concentration of the
reactant will decrease exponentially with time, and its half-life
will be constant. Does the fraction of molecules that react per
unit time change as the reaction progresses? Justify your
answer.
please justify with words and not just equations.

Consider the conversion of reactant A to product B. If
k1 is equal to 0.017 s-1, and if k2 is equal to
0.002 s-1, then, if the initial concentration of A, is 0.84 M, how
many seconds will it take to reach equilibrium? To make the problem
more tractable, assume that we are only considering the first-order
decrease in A until it reaches its equilibrium value. In other
words, ignore the back reaction. Report your answer to the nearest
second.

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