Question

Solve the following problems. The activation energy for the reaction below equals 1.0 x 105 J/mol. Given k 2.5 x 103 sec1 at 332 K, find k at 375 K. 1. N20s (g)-2NO (g) + 0a(g) Based on information in problem 1, find the temperature at which k is twice as large as it is at 332K. 2.
Please show all your work and write neatly please! Thank you guys! ??

Answer 1

1) ln (K₁ / K₂) =Ea / R [1/T₂ - 1/T₁]
K₁= 2.5*10^-3 sec^-1 ; T₁ = 332 K
K₂ = ? ; T₂ = 375 K
Ea = 1*10⁵ J/mol ; R = 8.314 J/mol K
ln (2.5*10^-3 / K₂) =1*10⁵ / 8.314 [1/375 - 1/332]
ln (2.5*10^-3 / K₂) = -4.154216
2.5*10^-3 / K₂ = e^-4.154216 = 0.015698
K₂ = 0.15925 sec^-1

2) T₁ = 332 K ; K₁ = 0.15925 sec^-1
T₂ = ? ; K₂ = 2*0.15925 = 0.3185 sec^-1
use the same formula
ln (0.15925 / 0.3185) = 1*10⁵ / 8.314 [1/T₂ - 1/332]
-5.7628*10^-5 = 1/T₂ - 1/332
T₂ = 338.4759 ~ 338.5 K