can someone help me with this question please thank you!!?
2017-02-26 10:24 pm
A reaction has a pre-exponential factor of A = 1700 dm3 mol-1 s-1 and an activation energy of 17.1 kJ mol-1. At which temperature will the reaction proceed twice as quickly as it does at 283.2 K? Give your answer to 1 decimal place and use R = 8.3145 J K-1 mol-1 and 0 °C = 273.15 K.
回答 (1)
2017-02-26 10:48 pm
Eₐ = 17.1 kJ mol⁻¹ = 17100 J mol⁻¹
R = 8.3145 J K⁻¹ mol⁻¹
Arrhenius equation : k = A e^[-Eₐ/(RT)]
i.e. ln(k₁/k₂) = -(Eₐ/R) [(1/T₁) - (1/T₂)]
ln(k₁/k₂) = -(17100/8.3145) [(1/T₁) - (1/T₂)]
The reaction at T₂ K proceeds twice as quickly as it does at T₁ = 283.2 K.
ln(1/2) = -(17100/8.3145) [(1/283.2) - (1/T₂)]
(1/283.2) - (1/T₂) = -8.3145 ln(1/2) / 17100
1/T₂ = (1/283.2) + [8.3145 ln(1/2) / 17100]
T₂ = 1 / {(1/283.2) + [8.3145 ln(1/2) / 17100]}
T₂ = 313.1 (K)
The temperature at which the reaction proceeds twice as quickly as it does at 283.2 K = 313.1 K
R = 8.3145 J K⁻¹ mol⁻¹
Arrhenius equation : k = A e^[-Eₐ/(RT)]
i.e. ln(k₁/k₂) = -(Eₐ/R) [(1/T₁) - (1/T₂)]
ln(k₁/k₂) = -(17100/8.3145) [(1/T₁) - (1/T₂)]
The reaction at T₂ K proceeds twice as quickly as it does at T₁ = 283.2 K.
ln(1/2) = -(17100/8.3145) [(1/283.2) - (1/T₂)]
(1/283.2) - (1/T₂) = -8.3145 ln(1/2) / 17100
1/T₂ = (1/283.2) + [8.3145 ln(1/2) / 17100]
T₂ = 1 / {(1/283.2) + [8.3145 ln(1/2) / 17100]}
T₂ = 313.1 (K)
The temperature at which the reaction proceeds twice as quickly as it does at 283.2 K = 313.1 K
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