aA → bB [A]_0 = 2.80×10^(-3) M +3.60×10^(-2) L/mol.s

A certain reaction has the following general form:

aA → bB

At a particular temperature and [A]_0 = 2.80×10^(-3) M,

concentration versus time data were collected for this reaction,

and a plot of 1/[A] versus time resulted in a straight line

with a slope value of +3.60×10^(-2) L/mol•s.

a. Determine the rate law, the integrated rate law,

and the value of the rate constant for this reaction.

b. Calculate the half-life for this reaction.

c. How much time is required for the concentration of A

to decrease to 7.00×10^(-4) M?

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[참고] 2차 반응

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a. Determine the rate law, the integrated rate law,

and the value of the rate constant for this reaction.

slope value = +3.60×10^(-2) L/mol•s

= value of the rate constant

기울기 = 속도 상수 이고,

속도 상수의 단위 = L/mol•s 이므로,

---> 2차 반응.

b. Calculate the half-life for this reaction.

2차 반응의 반감기

t_1/2 = 1 / (k [A]_0)

= 1 / [(+3.60×10^(-2) /M•s) × (2.80×10^(-3) M)]

= 1 / [(3.60×10^(-2)) × (2.80×10^(-3))]

= 9920.6 s

= 9.92×10^3 s

c. How much time is required for the concentration of A

to decrease to 7.00×10^(-4) M?

2차 반응 속도식

(1 / C_t) = k•t + (1 / C_0)

t = [(1 / C_t) – (1 / C_0)] / k

= [(1 / (7.00×10^(-4))) – (1 / (2.80×10^(-3)))] / (3.60×10^(-2))

= 29761.9 s

= 2.98×10^4 s

[키워드] 2차 반응 기준, second-order reaction graph