Given that the initial rate is .0160s-1 at an initial temperature of 29 degrees

Question

Given that the initial rate is .0160s-1 at an initial temperature of 29 degrees C, what would the rate constant be at a temperature of degrees C for the same reaction described in part A. ( T2= 42.9 C) given the activation energy of a certain reaction 39.5 kJ/mol. Given that the initial rate is .0160s-1 at an initial temperature of 29 degrees C, what would the rate constant be at a temperature of degrees C for the same reaction described in part A. ( T2= 42.9 C) given the activation energy of a certain reaction 39.5 kJ/mol. Given that the initial rate is .0160s-1 at an initial temperature of 29 degrees C, what would the rate constant be at a temperature of degrees C for the same reaction described in part A. ( T2= 42.9 C) given the activation energy of a certain reaction 39.5 kJ/mol.

Explanation / Answer

Solution :-

Initial rate constant K1 = 0.106 s-1

initial temperature T1 = 29 C +273 = 302 K

Final rate constant K2 = ?

Final temprature T2 = 42.9 C +273 = 315.9 K

Activation energy Ea = 39.5 kJ per mol * 1000 J / 1 kJ = 39500 J per mol

Now lets use the Arhenius equation to calculate the final rate constant

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

where R= 8.314 J per mol. K

lets put the values in the formula and calculate the activation energy

ln[K2/0.160] = (39500 J per mol / 8.314 J per K mol )*[(1/302)-(1/315.9)]

ln[K2/0.160] = 0.69222

K2/0.160]= anti ln [ 0.69222]

K2/0.160 = 1.998

K2 = 0.160 s-1 * 1.998

K2 = 0.320 s-1

Threfore the final rate constant K2 = 0.320 s-1

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