Part 1 A certain reaction with an activation energy of 115 kJ/mol was run at 555

Question

Part 1

A certain reaction with an activation energy of 115 kJ/mol was run at 555 K and again at 575 K . What is the ratio of f at the higher temperature to f at the lower temperature?

Part 2

The activation energy of a certain reaction is 30.8 kJ/mol . At 21  C , the rate constant is 0.0110s1. At what temperature in degrees Celsius would this reaction go twice as fast?

Part 3

Given that the initial rate constant is 0.0110s1 at an initial temperature of 21  C , what would the rate constant be at a temperature of 150.  Cfor the same reaction described in Part A?

Part A

A certain reaction has an activation energy of 68.0 kJ/mol and a frequency factor ofA1 = 7.60×1012 M1s1 . What is the rate constant, k, of this reaction at 29.0 C ?

Express your answer with the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot (asterisk) or a dash.

Part B

An unknown reaction was observed, and the following data were collected:

Determine the activation energy for this reaction.

Part C

The rate constant for this reaction is ________ s-1.

Part D

A second-order reaction has a rate law: Rate = k[A]2, where k = 0.150 M1s1. If the initial concentration of A is 0.250 M, what is the concentration of A after 5.00 minutes? Be sure to pay attention to the units in this problem so that they cancel out.

Explanation / Answer

1) As per Arrhenius Equation:

ln(k2/k1) = (Ea/R)*[(1/T1) - (1/T2)] ; where k1 & k2 are the rate constants at temperature T1 & T2 respectively

Thus, ln(k2/k1) = (115000/8.314)*[(1/555) - (1/575)] = 0.867

Thus, k2/k1 = 2.379

2) As per Arrhenius Equation:

ln(k2/k1) = (Ea/R)*[(1/T1) - (1/T2)] ; where k1 & k2 are the rate constants at temperature T1 & T2 respectively

Now, for the reaction to go twice as fast as early, (k2/k1) = 2

Thus, ln2 = (30800/8.314)*[(1/294) - (1/T2)]

or, T2 = 311.11 K = 38.11 0C

3) As per Arrhenius Equation:

ln(k2/k1) = (Ea/R)*[(1/T1) - (1/T2)] ; where k1 & k2 are the rate constants at temperature T1 & T2 respectively

Thus, ln(k2/0.011) = (115000/8.314)*[(1/294) - (1/423)]

or, ln(k2/0.011) = 13.1

or, (k2/0.011) = 489089

or, k2 = rate constant at 150 0C = 5380 s-1

A) As per Arrhenius Equation:-

lnk = lnA - (E/(RT))

or, lnk = ln(7.6*1012) - (68000/8.314*302) = 2.576

or, k = rate constant = 13.15 s-1

B) As per Arrhenius Equation:

ln(k2/k1) = (Ea/R)*[(1/T1) - (1/T2)] ; where k1 & k2 are the rate constants at temperature T1 & T2 respectively

Thus, ln(185/109) = (Ea/8.314)*[(1/352) - (1/426)]

or, Ea = Activation energy = 8912.36 J = 8.913 kJ

c) The governing equation for 1st order reaction is :-

k*t = ln(A0/A) ; where A0 & A are the concentration of the reactants at time t = 0 and time t = t respectively

Thus, k*20 = ln(0.2/0.05)

or, rate constant , k = 0.0693 s-1

d) The governing equation for 2nd order reaction is :-

k*t = (1/A) - (1/A0)

or, 0.15*300 = (1/A) - (1/0.25)

or, A = concentration after 5 minutes = 0.0204 M

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