each of five different t verses inverse temperature. Two points on this line are
ID: 703779 • Letter: E
Question
each of five different t verses inverse temperature. Two points on this line are stants never change.. unless you change the temperature of a reaction. N explores the temperature dependence of the reaction below, by finding k a emperatures. The student creates the following, plot of In k y 4.1 at x 0.0015 and y 2.2 at x-0.00163 23 23 3,5- 2.5 0.S 0.0017 0,00165 0,00155 0,0015 What is the activation energy (Ea) for this reaction? What is the pre exponential term tor this reaction (A)? RT What is k for this reaction, if it occurs at 660 K? Page 7 of 8Explanation / Answer
From the Arrhenius equation
At temperature T1
k1 = A exp(-Ea/RT1)
Take natural logarithm on both sides
ln k1 = ln A - Ea/RT1......... Eq1
At temperature T2
k2 = A exp(-Ea/RT2)
Take natural logarithm on both sides
ln k2 = ln A - Ea/RT2......... Eq2
Subtract the Eq2 from eq1
ln k1 - ln k2 = - Ea/RT1 + Ea/RT2
ln k1 - ln k2 = (Ea/R) (1/T2 - 1/T1)
Given that
x1 = 1/T1 = 0.0015
y1 = ln k1 = 4.1 x 10^-4
x2 = 1/T2 = 0.00165
y2 = ln k2 = 2.2 x 10^-4
Put the values
ln k1 - ln k2 = (Ea/R) (1/T2 - 1/T1)
4.1 x 10^-4 - 2.2 x 10^-4 = (Ea/8.314) (0.00165 - 0.0015)
0.00019 = 1.804 x 10^-5 x Ea
Ea = 10.53 J/mol
ln k1 = ln A - Ea/RT1
ln A = ln k1 + Ea/RT1
ln A = 4.1 x 10^-4 + 10.53*0.0015/8.314 = 0.00231
A = 1.0023 cm3 mol-1 s-1
At T3 = 600K
k3 =?
ln k1 - ln k3 = (Ea/R) (1/T3 - 1/T1)
4.1 x 10^-4 - ln k3 = (10.53/8.314) (1/600 - 0.0015)
4.1 x 10^-4 - ln k3 = 0.0002111
ln k3 = 0.0001989
k3 = 1.0001989 cm3 mol-1 s-1
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