The rale constant for a gas phase bimolecular elementary reaction is given as be

ID: 1073639 • Letter: T

Question

The rale constant for a gas phase bimolecular elementary reaction is given as below. Discuss the rate constant above with collision theory associated with successful collisions. K_r = P sigma (8KT/pi mu)^1/2 N_A e^-A alpha/RT Here P: steric factor, o: collision cross-sectional area, mu: reduced mass of u two-particle system The Arrhenius equation is k_r = Ae^-Ea/RT or lnk_r = InA - E_a/RT. The rate of the second-order decomposition of acetaldehyde (ethanol, CH_3CHO) was measured over the temperature range 700-1000K, and the rate constants are provided below. Determine A and E_a. Provide a molecular interpretation for the observation that the viscosity of a gas increases with temperature whereas the viscosity of a liquid decreases with increasing temperature. What is the rate-determining step? Is the slow step rate-determining step in the following figure? Why or why not?

Explanation / Answer

2. Given kr = Ae^(-Ea/RT) and ln kr = ln A – Ea/RT….(1)

The equation is of the form y = mx + c where m = slope = -Ea and c = ln A.

Ea and A can be found from a plot of ln kr vs 1/T. Prepare the table as below:

T (K)

1/T (K-1)

kr (dm3 mol-1 s-1)

ln kr

300

0.00333

7.9*106

15.882

350

0.00286

3.0*107

17.217

400

0.00250

7.9*107

18.185

450

0.00222

1.7*108

18.951

500

0.00200

3.2*108

19.584

Plot a graph of ln kr vs 1/T as shown below:

The graph is plotted above. The slope of the plot is -2769.5. Therefore,

-2769.5 = -Ea

===> Ea = 2769.5

Since Ea must have the unit of J/mol, hence the activation energy for the reaction is Ea = 2769.5 J/mol = 2.7695 kJ/mol 2.77 kJ/mol (ans).

From the plot, c = 25.115; therefore, ln A = 25.115

====> A = e^(25.115) = 8.078*1010

Since A must have the unit of kr (the exponential term is dimensionless), we must have A = 8.078*1010 dm3 mol-1 s-1 (ans).

4. The slow step is the rate-determining step of the reaction. The slow step is the rate-determining step because the entire reaction has to wait for the slow step to get over and hence the rate of the reaction must depend on the rate of the slow step.