Question

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)

Answer 1

Answer:

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.