Chemistry 102: Kinetics: rate constant

Question

The rate constant for the reaction below was determined to be 3.241×10^-5  s^–1 at 800 K. The activation energy of the reaction is 265  kJ/mol. What would be the value of the rate constant at 9.50×10²  K?

N2​O(g)N2​(g)+O(g)

Answer 1

ln(k2/k1) = -Ea/R (1/T2 - 1/T1) ... Arrhenius Equation

k1 = 3.24x10^-5

k2 =?

Ea = 265 kJ/mol

R = 8.314 J/molK = 0.008314 kJ/molK

T1 = 800K

T2 = 950K

Plugging in the values and solving for k2, we have...

ln (k2/3.24x10^-5) = -265/0.008314 (1/950 - 1/800)

ln (k2/3.24x10^-5) = -31,874 (0.00105 - 0.00125)

ln (k2/3.24x10^-5) = -31,874 (-0.0002)

ln (k2/3.24x10^-5) = +6.375

k2/3.24x10^-5 = 587

k2 = 1.90x10^-2 s^-1

(be sure to check the math)