Question

2. A reaction was studied at the temperatures listed in the table below, and at each temperature the value of the first-order rate coefficient was determined. Fit the Arrhenius expression, equation (1), and the collision theory rate coefficient expression, equation (2) to the data in the table. Qualitatively compare the accuracy of the two results. k=k, EXPRF k=k,VT exp() T(°F) k(s) 76.7 1.12 x 10-5 92.9 3.63 x 105 114.5 1.15 x 104 132.5 3.26 x 104 148.7 8.92 x 10"

Answer 1

In the given equations temperature must be in Kelvin. So (^oF) values are converted to K.

Equation 1:

$k = k_{o}e^{-\frac{E}{RT}}$

$\Rightarrow ln\, k = ln\, k{_{o}}-\frac{E}{RT}$

This equation shows a straight line graph of ln k versus 1/T

with slope of (-E/R)

Equation 2:

$k = k_{o}\sqrt{T}e^{-\frac{E}{RT}}$

$\Rightarrow ln\, k = ln\, k{_{o}}+ln(\sqrt{T} )-\frac{E}{RT}$

$\Rightarrow ln\, (\frac{k}{\sqrt{T}}) = ln\, k{_{o}}-\frac{E}{RT}$

This equation shows a straight line graph of ln (k/T^0.5) versus 1/T

with slope of (-E/R)

0.0029 0.003 0.0031 0.0032 0.0033 0.0034 Arrhenius y = -10790x + 24.83 R2 = 0.998 Collision Ink and In(k/T^0.5) y = -10632x+ 21.45 R2 = 0.998 1/T

Both the graphs fit well (R² = 0.998) with the data provided. But Collision theory shows requirement of lower activation energy than Arrhenius theory (slope of equation 2 is less than equation 1)