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A second-order reaction has a rate constant of 0.009500/(M · s) at 30°C. At 40°C, the...

A second-order reaction has a rate constant of 0.009500/(M · s) at 30°C. At 40°C, the rate constant is 0.02300/(M · s).

(A) What is the activation energy for this reaction? ______kJ/mol

(B) What is the frequency factor, A? _______ /(M*s)

(C) Predict the value of the rate constant at 50°C. ________ /(M*s)

Solutions

Expert Solution

A second-order reaction has a rate constant of 0.009500/(M · s) at 30°C. At 40°C, the rate constant is 0.02300/(M · s).

(A) What is the activation energy for this reaction? ______kJ/mol

Solution :- K1 = 0.009500 M-1 s-1

K2 = 0.02300 M-1 s-1

T1 = 30 C +273 = 303 K

T2 = 40 C +273 = 313 K

Ea = ?

Formual

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

ln [ 0.02300/0.009500] = Ea / 8.314 J per mol K [(1/303)-(1/313)]

0.8842 = [Ea / 8.314 J per mol K] * 0.00010544

0.8842 * 8.314 J per mol K / 0.00010544 K = Ea

69719 J per mol = Ea

Lets convert it to kJ

69719 J per mol * 1 kJ /1000 J = 69.7 kJ/mol

So the activation energy of the reaction is 69.7 kJ /mol

(B) What is the frequency factor, A? _______ /(M*s)

Solution :- formula

K= A e^(-Ea/RT)

Lets use the T1 and K1 and find the frequency factor

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

   = 0.0095 / e^(-69719 / 8.314 * 303)

= 9.93*10^9

So the frequency factor is 9.93*10^9

(C) Predict the value of the rate constant at 50°C. ________ /(M*s)

Solution :-

K1 = 0.009500 M-1 s-1

T1 = 303 K

K2 = ?

T2 = 50 C +273 = 323 K

Ea = 69719 J per mol

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

ln [ K2/0.009500] = 69719 J per mol / 8.314 J per mol K [(1/303)-(1/323)]

ln [ K2/0.009500] = 1.714

K2/0.009500 = anti ln [1.714]

K2/0.009500 = 5.55

K2 = 5.55 * 0.009500 M-1 s-1

K2 = 0.0527 M-1 s-1

So the rate constant at 50 C is 0.0527 M-1 s-1

queen_honey_blossom answered 2 years ago
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