A certain reaction has an activation energy of 63.0 kJ/mol and a frequency facto

Question

A certain reaction has an activation energy of 63.0 kJ/mol and a frequency factor of A1 = 6.50 Times 1012 M-1 s-1. What is the rate constant, k, of this reaction at 24.0 degree C? Express your answer with the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot (asterisk) or a dash. The linear form of the Arrhenius equation The Arrhenius equation can be rearranged to a form that resembles the equation for a straight line: where y is ln k, m is the slope or - Es/R, x is 1/T, and b is the y-intercept or ln A. The linearity of this equation is illustrated graphically in the image. An unknown reaction was observed, and the following data were collected: Determine the activation energy for this reaction.

Explanation / Answer

Solution :-

Part A)Formula to calculate the rate constant using the frequency factor and activation energy is

K = A*e^(-Ea/RT)

A = frequency factor

R= 8.314 J per mol K , T= Kelvin temperature

Ea= 63.0 kJ per mol * 1000 J/1kJ = 63000 J peer mol

R= 24 C +273 = 297 K

A= 6.50*10^12 M-1s-1

Lets put the values in the formula

K= 6.50*10^12 M-1s-1* e^(-63000 J per mol / 8.314 J per mol K * 297 K)

K= 54.0 M-1s-1

Hence rate constant = 54.0 M-1.s-1

Part B) using the Arrhenius equation we can calculate the activation energy

T1 = 352 K and K1 = 109

T2 = 426 K and K2=185

Arrhenius equation is as follows

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

lets put the values in the formula

ln[185/109]=Ea/ 8.314 J per mol K * [(1/352)-(1/426)]

0.529 = Ea/ 8.314 J per mol K * 0.0004935 K

Ea = 0.529 * 8.314 J per mol K / 0.0004935 K

Ea= 8912 J per mol

Lets convert it to the kJ

8912 J per mol * 1 kJ / 1000 J = 8.91 kJ per mol

Therefore the activation energy for the reaction is 8.91 kJ per mol

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