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 of A1 = 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

Part 1 answer ln(k2/k2) = (145 kJ/mol / 8.3144x10^-3) x (1/545-1/565)
ln(k2/k1) = 17440 x (6.495x10^-5) = 1.1327
k2/k1 = 3.10 times faster at 565 than at 545.

Part 2 answer

ln(k/k) = - (Ea/R)[(1/T) - (1/T)]

Let T = 293 K, so k = 0.01301/s. Also, we know Ea = 48.8 kJ/mol

Set –[(1/T) - (1/T)] = (R/Ea)ln(k/k) and plug in for all the knowns:

–(1/T) + (1/293 K) = [(8.314 J/mol•K)/(48800 J/mol)]ln(2)

(1/T) = [(1/293) – 1.18 × 10] K¹ = 3.2949 × 10³

Part 3 answer:

For both problems, you will use the Arrhenius equation which can be written as:

ln (k2/k1) = -Ea/R(1/T2-1/T1)

For the first one, k2 = 2Xk1, so k2/k1 = 2. You must convert your activation energy into J/mol, and express your temperatures in Kelvin. R will be 8.314 J/molK. It just becomes an algebra problem to solve for T2.

For the second question, you will use the same equation, except that you will be solving for k2.

T= 303.5 K

Note: You don't need to know the rate constant to determine the temperature at which the rate doubles - that is determined by the energy of activation. All that was needed to get that was the ratio of the rates - and that was given in the form of the question. If they had asked what the rate constant was after it doubled, we'd need the rate canstant at 293 K, before it doubled, obviously.

Part A answer,

Arrhenius Equation:
k = Ae^(-Ea/RT);
k = (4.20×10^12)e^((-67,000 J/mol)/(8.314 J/K·mol)(297.15 K));
k = 7.01

Part -B answer

Begin with the Arrhenius equation:

k = Ae^(-Ea/RT)

Take the ln of both sides

ln(k) = ln(A) - Ea/RT

Rearranging you get a y = mx + b form:
ln(k) = (-Ea/R) * (1/T) + ln(A)

y = ln(k) m = -Ea/R x = 1/T b = lnA

To find the activation energy we find the slope of the line given from the data points and multiply by -R.

slope = change in y / change in x : change in ln(k) / change in 1/T

[[ln(185)-ln(109)] / [(1/426 K - 1/352 K) ] * -8.314J/mol-K= 8.91 kJ/mol.

Part-C answer.

Examining the data you can find that reactant concentration halves every 10 seconds, i.e. your reaction has a constant half life period of
t½ = 10s

Because t½ is constant (=independent from concentration) reaction it must be a first order reaction.
Half life and rate constant for a first order reaction are related as:
k = ln(2)/t½

Hence,
k = ln(2) / 10s = 0.0693s¹ = 6.93×10² s¹

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