**k = Ae ^{-(Ea/RT)}**

Firstly, using the following experimental data for the rate constants and temperatures draw a graph of ln(k/s^{-1}) versus (1/T) x (10^{3} K) the reciprocal of the pure number obtained by dividing the temperature by its unit kelvin. Secondly, determine the values of A and E_{a} for this reaction. Thirdly, use the equation of the line from the graph to calculate the rate constants at 283 K (do not extrapolate the graph).

Here,

**Arrhenius plot temperature data.**

In above Figure the Arrhenius plot is a straight line. From a full-sized manually-drawn version of Figure I obtained y = -9.98x + 28.7. The Arrhenius equation has a gradient m = -E,/R where R is the gas constant and E_{a} is the activation energy. So from the gradient we obtain the activation energy E_{a} we must also remember to indude the units and their multiples of the axes in the calculation.

The equation of the straight line in above figure has an intercept of c = 28.7 which for the Arrhenius equation is equal to c = In(A). The intercept is on the y-axis and so the intercept will have the same units and any multiples there happens to be as the y-axis.

**c = ln (A/s ^{-1}) = 28.7; A/s^{-1} = 2.91 x 10^{12}**

**A = 2.91 x 10 ^{12} s^{-1}**

To calculate the rate constant at: T= 283 K then (1/7) = (3.5336 x 10^{-3} K^{-1}) and so x = (3.5336 x 10^{-3} K^{-1}) using the equation for the best line we can then calculate the corresponding value for y = ln (k/s^{-1}).

**In (k/s ^{-1}) = (-9.98 x 10^{3} K) (3.5336 x 10^{-3} K^{-1}) + 28.7 = — 6.37 k**

**K = 1.40 x 10 ^{-3} s^{-1}**