Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 °C? (HINT: the activation energy for this reaction is 160 kJ/mol)

T (K) 1/T (K-1) ln k

462.9 2.160*10^ (-3) - 10.589

472.1 2.228*10^ (-3) - 9.855

503.5 1.986*10^ (-3) - 7.370

524.4 1.907*10^ (-3) - 5.757

A) 8.1x10^ (-15) s^ (-1)

B) 2.2x10^ (-13) s^ (-1)

C) 2.7x10^ (-9) s^ (-1)

D) 2.3x10^ (-1) s^ (-1)

E) 9.2x10^3 s^ (-1)

D) 2.3 x 10⁻¹ s⁻¹

Explanation:

The rate constant is related to the activation energy through the formula:

k = Ae^ (-Eₐ / RT)

where A is the collision factor, Eₐ the activation energy, R is the gas constant (8.314 J/Kmol), and T is the temperature (K)

So a plot of lnk versus 1/T (Arrehenius plot) gives us a straight line with slope equal - Eₐ/R and intercept lnA

lnk = - (Eₐ/T) (1/T) + lnA

which has the form y = mx + b

In this problem, we can use the data provided to:

a) Using a calculator determine the slope and intercept and then calculate the value of rate constant at 320 ºC, or

b) Plot the data and determine the equation of the best line, and answer the question for k @ 320 ºC by reading the value from the plot.

Once you do the plot, the resulting equation is:

y = - 19 x 10³ x + 30,582 (R² = 0.999)

So for T = 320 + 273 K = 593 K

Y = 19 x 10³ X + 30.58

So for T = (320 + 273) K = 593 K

Y = - 19 x 10³ (1/593) + 30.58 = - 32.04 + 30.58 = - 1.46

and then since

y = lnk ⇒ e^y = k

k = e^-1.46 = 2.3 x 10⁻¹ s⁻¹

Note: there is an error of transcription in the value for T = 472.1 (1/T = 2.118 x 10⁻³ and not 2.228 x 10⁻³). You can recognize this mistake if you plot the data and notice it produces an outlier.