Question

1.) The reaction

C4H8(g)⟶2C2H4(g)

has an activation energy of 262 kJ/mol. At 600.0 K the rate constant is 6.1×10−8 s−1. What is the value of the rate constant at 860.0 K?

?=_____ s−1

2.)

A certain reaction has an activation energy of 47.01 kJ/mol. At what Kelvin temperature will the reaction proceed 7.50 times faster than it did at 357 K?

____ K

3.) Consider this reaction data.

A⟶products

T (K)	k (s–1)
275	0.383
875	0.659

If you were going to graphically determine the activation energy of this reaction, what points would you plot? To avoid rounding errors, use at least three significant figures in all values.What is the activation energy of this reaction? Determine the rise, run, and slope of the line formed by these points.

point 1: ?=

point 1: ?=

Answer 1

1)

Given:

T1 = 600 K

T2 = 860 K

K1 = 6.1*10^-8 s-1

Ea = 262 KJ/mol

= 262000 J/mol

use:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(K2/6.1*10^-8) = (262000.0/8.314)*(1/600 - 1/860.0)

ln(K2/6.1*10^-8) = 31513*(5.039*10^-4)

K2 = 0.4801 s-1

Answer: 0.480 s-1

2)

Given:

T1 = 357 K

K2/K1 = 7.5/1

Ea = 47.01 KJ/mol

= 47010 J/mol

use:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(7.5/1) = (47010.0/8.314)*(1/357 - 1/T2)

2.0149 = 5654.318*(1/357 - 1/T2)

T2 = 409 K

Answer: 409 K

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