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일반화학/[14장] 화학반응 속도론

500℃ gas cyclopropane propene rate 6.7×10^(-4) s-1

by 영원파란 2022. 11. 21.
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500℃ gas cyclopropane propene rate 6.7×10^(-4) s-1

 

 

The conversion of cyclopropane to propene

in the gas phase is a first-order reaction and

has a rate constant at 500℃ of 6.7×10^(-4) s^-1.

a) If the initial concentration of cyclopropane was 0.25 M,

what is the concentration after 8.8 min?

b) How long will it take for the concentration of cyclopropane

to decrease from 0.25 M to 0.15 M?

c) How long will it take to convert 74% of the starting material?

 

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1차 반응 속도식

ln(C_t / C_0) = –k•t

( 참고 https://ywpop.tistory.com/26 )

 

 

 

a) If the initial concentration of cyclopropane was 0.25 M,

what is the concentration after 8.8 min?

 

ln(C_t / 0.25 M) = –(6.7×10^(-4) /s) (8.8×60 s)

 

C_t / 0.25 = e^[–(6.7×10^(-4)) (8.8×60)]

 

C_t = e^[–(6.7×10^(-4)) (8.8×60)] × 0.25

= 0.1755 M

= 0.18 M

 

 

 

b) How long will it take for the concentration of cyclopropane

to decrease from 0.25 M to 0.15 M?

 

ln(C_t / C_0) = –k•t

 

t = –ln(C_t / C_0) / k

= –ln(0.15 M / 0.25 M) / (6.7×10^(-4) /s)

= –ln(0.15 / 0.25) / (6.7×10^(-4))

= 762.43 s

 

762.43 / 60 = 12.707 min

= 13 min

 

 

 

c) How long will it take to convert 74% of the starting material?

 

초기 농도(양) = 100 이라 가정하면,

74% 전환되면,

100 – 74 = 26%

---> 남아있는 농도(양) = 26

 

t = –ln(C_t / C_0) / k

= –ln(26 / 100) / (6.7×10^(-4))

= 2010.56 s

 

2010.56 / 60 = 33.509 min

= 34 min

 

 

 

 

[참고] 1차 반응의 반감기

t_1/2 = 0.693 / k

---> 반감기 식에 농도 항이 없으므로,

---> 1차 반응의 반감기는 반응물의 초기 농도와 무관.

 

t_1/2 = 0.693 / (6.7×10^(-4) /s)

= 0.693 / (6.7×10^(-4))

= 1034.33 s

 

1034.33 / 60 = 17.239 min

= 17 min

 

 

 

 

Q. 초기 농도 0.50 M, 12분 후 농도

C_t = e^[–(6.7×10^(-4)) (12×60)] × 0.50

= 0.30865 M

= 0.31 M

 

 

 

Q. 0.30 M에서 0.15 M로 감소하는 시간 (= 반감기)

t = –ln(0.15 / 0.30) / (6.7×10^(-4))

= 1034.548 s

 

1034.548 / 60 = 17.2425 min

= 17 min

 

 

 

 

[키워드] 1차 반응 기준문서, 일차 반응 기준문서

 

 

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