uspective temperatures we can determine the Ea ll ellergy an

uspective temperatures, we can determine the Ea ll ellergy) and A (frequency factor) by plotting In(kyversus T-1, The slope of the straight line equals and the y-intercept equal IncA). Knowing Ea and k at one temperature, we can calculate k at another temperature. variable. .In general, knowing 4 out of 5 variables (k, T, ka, Ta, En), we can solve for the f or the fifth Solve the following problems. The activation energy for the reaction below equals 1.0 x 105 J/mol. Given k = 2.5 x 10-3 sec-1 at 332 K, find k at 375 K. 1. N20s (g) 2NO2(g) + 1/2 O2(g) Based on information in problem 1, find the temperature at which k is twice as large as it is at 332K. 2. Di-tert-butylperoxide (DTBP) is used as a catalyst in making polymers. In the gaseous state, DTBP decomposes to acetone and ethane by a first order reaction 3.

Solution

1) Arrhenius equation

ln(K2/k1) = Ea/R[1/T1 - 1/T2]

k1 = 2.5*10^-3 s-1 , T1 = 332 k

k2 = x s-1 , T2 = 375 k

Ea = 1.0*10^5 kj/mol , R = 8.314*10^-3 j.k-1.mol-1

ln(x/(2.5*10^-3)) = ((1*10^5)/(8.314))((1/332)-(1/375)

K2 =rate constant at 375 k = 0.16 s-1

2)

Arrhenius equation

ln(K2/k1) = Ea/R[1/T1 - 1/T2]

k1 = 2.5*10^-3 s-1 , T1 = 332 k

k2 = 2*2.5*10^-3 s-1 , T2 = x k

Ea = 1.0*10^5 kj/mol , R = 8.314*10^-3 j.k-1.mol-1

ln((2*2.5*10^-3)/(2.5*10^-3)) = ((1*10^5)/(8.314))((1/332)-(1/x)

x = 338.5

T2 = 338.5 k , the rate constant is doubled.