1.For the decomposition of Ni(CO)4 in the gas phase, these rate constants were o

ID: 704540 • Letter: 1

Question

1.For the decomposition of Ni(CO)4 in the gas phase, these rate constants were obtained experimentally.

T(? C)                                            k(s-1)
47.3                                              0.233
50.9                                              0.354
55.0                                              0.606
60.0                                              1.022
66.0                                              1.873

Plot the data using the linearized Arrhenius equation. Fully label the graph, axes, data and line fit. Include the regression analysis, and a discussion of the quality of the regression analysis.
a). Calculate the activation energy and pre-exponential factor for this reaction.
b). Estimate the rate constant of the reaction at 40.0 ?C.

Explanation / Answer

The integrated form of the Arrhenius equation is given as

ln k = -Ea/RT + ln A

where k is the rate constant of the reaction at absolute temperature T. Ea is the activation energy of the reaction and A is the pre-exponential factor (also known as frequency factor).

Prepare a table of 1/T vs ln k as below.

t (°C)

T (K) = (t + 273)

1/T (K-1)

k (s-1)

ln k

47.3

320.3

0.003122

0.233

-1.4567

50.9

323.9

0.003087

0.354

-1.0384

55.0

328.0

0.003049

0.606

-0.5009

60.0

333.0

0.003003

1.022

0.0218

66.0

339.0

0.002950

1.873

0.6275

Plot ln k vs 1/T as below.

Plot of ln k vs 1/T

The regression equation is shown on the plot. The R2 value is close to 1.0000 indicating that the data points fit the curve well.

a) Compare the integrated rate law with the regression equation. We have

-Ea/R = -12174

This is the slope of the plot and will have unit K. We find the unit by dimension analysis. The left hand side is dimensionless while 1/T has unit K-1. To make the right hand side dimensionless, the slope must have unit K. Therefore,

Ea/R = 12174 K

====> Ea = (12174 K)*R

====> Ea = (12174 K)*(8.314 J/mol.K) =101214.636 J/mol

====> Ea = (101214.636 J/mol)*(1 kJ/1000 J) = 101.214636 kJ/mol

====> Ea ? 101 kJ/mol (ans).

Comparing the two equations again, we have,

ln A = 36.566

=====> A = exp(36.566) = 7.5929*1015 ? 7.59*1015

A will have the same unit as k; hence, the pre-exponential factor is 7.59*1015 s-1 (ans).

b) We have t = 40.0°C; therefore, T = (40.0 + 273) K = 313.0 K and x = 1/T = 1/(313.0 K) = 0.003195 K-1. Put x = 0.003195 K-1 in the regression equation and get

y = -(12174 K)*(0.003195 K-1) + 36.566

====> y = -38.89593 + 36.566 = -2.32993

We know that y = ln k; therefore,

ln k = -2.32993

====> k = exp(-2.32993) = 0.09730 ? 0.097

k has unit s-1; hence, the rate constant of the reaction at 40.0°C is 0.097 s-1 (ans).