A certain reaction has an activation energy of 28.39 kJ/mol. At what Kelvin temp

Question

A certain reaction has an activation energy of 28.39 kJ/mol. At what Kelvin temperature will the reaction proceed 3.50 times faster than it did at 299 K?

A certain reaction has an activation energy of 28.39 kJ/mol. At what Kelvin temperature will the reaction proceed 3.50 times faster than it did at 299 K?

Res o 3/26/2016 11:55 PM @ 30.1/503/24/2016 02:32 PM Gradebook Calculator-d Periodic Table lator Periodic Table Question 4 of 18 Map sapling learning ertain reaction has an activation energy of 28.39 kJ/mol. At what Kelvin temperature will the reaction proceed 3.50 times faster than it did at 299 K? Number Previous e Check Answer Next Exit- Hint 114 about us careers partners privacy policy terms of use contact us |he

Explanation / Answer

We must plot via ---> Ahrrenius Equation for 1 and several Points

According to Arrhenius, we can relate the rate constants as following:

K = A*exp(-Ea/(RT))

Where:

K = rate constant at Temperature “T”

A = Frequency Factor

E = Activation Energy in J/mol

R = ideal gas constant, 8.314 J/mol-K

T = absolute temperature

Ahrrenius Equation 2 Points

From Ahrrenius equation;

K1 = A*exp(-Ea/(RT1))

K2 = A*exp(-Ea/(RT2))

Note that A and Ea are the same, they do not depend on Temperature ( in the range fo temperature given)

Then

Divide 2 and 1

K2/K1 = A/A*exp(-Ea/(RT2)) / exp(-Ea/(RT1))

Linearize:

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

get rid of negative sign

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

k2 = 3.5*k1

Ea = 28390 J/mol

T1 = 299; T2 = T2; R= 8.314

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

ln(3.5*K1/K1) = 28390 /8.314*(1/299-1/T2)

solve for T2

ln(3.5) * 8.314 / 28390 - 1/299 = -1/T2

-0.0029776 = -1/T2

T2 = 1/0.0029776

T2 = 335.84 K

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