아레니우스 식. 565℃ 0.066 L/mol•min 728℃ 22.8 L/mol•min

아레니우스 식. 565℃ 0.066 L/mol•min 728℃ 22.8 L/mol•min

 

 

다음 반응의 속도 상수는 565℃에서 0.066 L/mol•min이고,

728℃에서는 22.8 L/mol•min이다.

 

 

For the reaction: 2N2O(g) → 2N2(g) + O2(g)

the rate constant is 0.066 L/mol•min at 565℃ and 22.8 L/mol•min at 728℃.

a) What is the activation energy(in kJ/mol) for this reaction?

b) What is k at 485℃?

c) At what temperature is k equal to 11.6 L/mol•min?

 

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a) What is the activation energy(in kJ/mol) for this reaction?

 

> k_1 = 0.066 L/mol•min, T_1 = 273.15 + 565 = 838.15 K

> k_2 = 22.8 L/mol•min, T_2 = 273.15 + 728 = 1001.15 K

 

 

 

ln(k_2 / k_1) = Ea/R (1/T_1 – 1/T_2)

( R = 8.314 J/K•mol )

( 참고 https://ywpop.tistory.com/7288 )

 

Ea = [ln((k_2) / (k_1)) × 8.314] / [(1/T_1) - (1/T_2)]

= [ln(22.8 / 0.066) × 8.314] / [(1/838.15) - (1/1001.15)]

= 250159.8 J/mol

= 250. kJ/mol

 

 

 

 

b) What is k at 485℃?

 

> k_1 = 0.066 L/mol•min, T_1 = 273.15 + 565 = 838.15 K

> k_2 = ? L/mol•min, T_2 = 273.15 + 485 = 758.15 K

 

 

 

ln(k_2 / k_1) = Ea/R (1/T_1 – 1/T_2)

ln(? / 0.066) = (250159.8 / 8.314) ((1/838.15) – (1/758.15))

 

? = e^[(250159.8 / 8.314) ((1/838.15) – (1/758.15))] × 0.066

= 0.001494 L/mol•min

 

 

 

 

c) At what temperature is k equal to 11.6 L/mol•min?

 

> k_1 = 0.066 L/mol•min, T_1 = 273.15 + 565 = 838.15 K

> k_2 = 11.6 L/mol•min, T_2 = ? K

 

 

 

ln(k_2 / k_1) = Ea/R (1/T_1 – 1/T_2)

ln(11.6 / 0.066) = (250159.8 / 8.314) ((1/838.15) – (1/?))

 

(1/838.15) – (1/?) = ln(11.6 / 0.066) / (250159.8 / 8.314)

1/? = (1/838.15) - (ln(11.6 / 0.066) / (250159.8 / 8.314))

 

? = 1 / [(1/838.15) - (ln(11.6 / 0.066) / (250159.8 / 8.314))]

= 979.13 K

 

 

 

979.13 – 273.15 = 705.98 = 706℃