# CHEM126 – CHAPTER 13 – REGULAR HOMEWORK / KINETICS II

## CHAPTER 13 REGULAR HOMEWORK

### Question 1

Initial rate of reaction data is given in the next question for the following reaction: 2A + B → C +D The rate of reaction = k[A]x[B]y. What are x and y, that is, what is the order of reaction with respect to A and B?

Select one or more:

### Explanation

You’ve done this before! Go for it!

### Question 2

Determine the rate constant from the concentration-time dependence (The units for the initial rate of reaction are Ms-1):

 [A] [B] Init. Rate 0.020 0.15 0.055 0.035 0.15 0.096 0.050 0.35 0.75

### Explanation

After finding x and y in Q1, use the rate equation: to solve for k.

### Question 3

Consider the second-order decomposition of nitroysl chloride:
2NOCl(g) → 2NO(g) + Cl2(g) At 450 K the rate constant is 15.4 atm-1s-1. How much time (in s) is needed for NOCl originally at a partial pressure of 47 torr to decay to 29.5 torr?

### Explanation

1. Convert torr to atm (this is done because our k value is in atm)
2. Treat the atm values as you would concentration and use the second order integrated rate law to solve for t (seconds needed)

### Question 4

How much time (in s) is needed for NOCl originally at a concentration of 0.0184 M to decay to 0.0027 M?

### Explanation

1. Convert M values to atm (this is done since our k value is in atm):
2. Use second order integrated rate law with atm values instead of concentrations to solve for t

### Question 5

The decomposition of formic acid (see below) is measured at several temperatures. HCOOH(g) → CO2(g) +H2(g) The temperature dependence of the first-order rate constant is: T(K) . . . . k(s-1)

```800 ....0.00027
825 ....0.00049
850 ....0.00086
875 ....0.00143
900 ....0.00234
925 ....0.00372```

Calculate the activation energy, in kJ/mol. Use all data points and do a linear regression using calculator or Excel. Do not pick 2 data points. This is less accurate and assumes all data points are equally good.

### Explanation

See the same type of problem done on Khan Academy here. For reference, here is what your plot will look like:

### Question 6

Calculate the pre-exponential term (in s -1).

### Explanation

Remember the form of the Arrhenius equation that resembles a line (y=mx+b)? You should have used it in Q5:
Notice how lnA is equivalent to b (the y-intercept). Take the y-intercept of the line that you plotted in Q5 and set it equal to lnA to solve for A.

### Question 7

Yikes! Looks like this one doesn’t have an answer. You’re on your own!

### Explanation

Make sure your units are consistent and use K.

### Question 8

A catalyst decreases the activation energy of a particular exothermic reaction by 29 kJ/mol, to 28 kJ/mol. Assuming that the mechanism has only one step, and that the products are 35 kJ lower in energy than the reactants, sketch approximate energy-level diagrams for the catalyzed and uncatalyzed reactions. What is the activation energy for the uncatalyzed reverse reaction?

### Explanation

The problem is easy if you follow instructions and draw a sketch.

### Question 9

Select one or more:

### Explanation

In this one-step mechanism, any change in the reactant concentrations will cause a change in the reactant rate. Since the coefficients of the reactants are 1, their orders are also 1. Note, the coefficients equal the reactant orders in one step equations. This may not be the case for equations with several step mechanisms.

### Question 10

Select one or more:

### Explanation

In this case, the slow step is the rate-determining reaction and is equal to the overall reaction rate. Any changes in reactant concentration in the fast step do not have a significant effect on the overall reaction rate. Changing [NO] will not affect reaction rate because it is only found in the fast step. Thus, its order is 0. Changing [O3[ will affect reaction rate because it is found in the slow step. It also has a coefficient of one. Therefore, its order is 1. Algebraic explanation:
See here for a more in depth explanation of this concept.

### Question 11

Select one or more:

### Explanation

N2O4 is the intermediate because it is created and consumed within the reaction-is not a reactant nor a product. This question is slightly different than the previous one in that the slow step depends on the products of the fast step to occur–an important distinction.

• N2O4 is made from two molecules of NO2 (in the reversible fast reaction) and so its concentration will depend on the concentration of NO2.
• That means that you can replace [N2O4] by [2NO2]
• Our initial rate equation for the slow reaction was:
• Our new rate equation (with replacement) is:
The NO2’s order of 2 is due to the coefficient of NO2.
See here for a more in depth explanation of this concept.

### Question 12

Select one or more:

Same as above.

### Question 13

What is the reaction order at each temperature?
Select one or more:

### Explanation

reaction order not dependent on temp. Stays the same. Rate constant is dependent Plot. ln[A] v. time gives straight line–first order

### Question 14

What is the rate constant at 310 K? (Use SI units. For example, use seconds not minutes or hours.)

### Explanation

1st order int. rate law

### Question 15

What is the rate constant at 315K?

### Explanation

1st order int. rate law

### Question 16

What is the activation energy (in J/mol)?

### Explanation

Use the Arrhenius equation.

### Question 17

When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay: HCOOH(g) →CO2(g) + H2 (g) The rate of reaction is monitored by measuring the total pressure in the reaction container. Time (s) . . . P (torr) 0 . . . . . . . . . 220 50 . . . . . . . . 324 100 . . . . . . . 379 150 . . . . . . . 408 200 . . . . . . . 423 250 . . . . . . . 431 300 . . . . . . . 435 At the start of the reaction (time = 0), only formic acid is present. What is the formic acid pressure (in torr) when the total pressure is 406? Hint: use Dalton’s law of partial pressure and the reaction stoichiometry.

### Explanation

HCOOH –> CO2 + H2 220 0 0 (t = 0) – x + x + x (change) 220 – x x x (total sum is total pressure = 406) 220 – x + x + x = 406 220 + x = 406 x = 186 Solve for 220 – x = 220 – 186 = 34 torr.

### Question 18

What is the half-life (in s)?

### Explanation

1) Do an ICE table to find partial pressure of HCOOH at 50s: HCOOH–> CO2 + H2 220 torr 0 torr 0 torr -x +x +x 220-x +2x =324 220+x=324 x=104 since x=the amount of reactant used up at 50s, 220-104= 116 torr HCOOH at 50s 2) Use first order integrated rate law to find k: ln(220/116)=-kt ln(220/116)=-k(50) k=1.28 x 10^-2 3) Use first order half life equation to find half life: t½ = 0.693 / k t½ =0.693/(1.28×10^-2) t½=54.14

### Question 19

What is the rate constant (in s-1)? (To determine the rate, you must use a reactant concentration or pressure. Use the procedure of in the previous question to determine the HCOOH pressure as a function of the total pressure.)

### Explanation

You have already found this in the previous Q. If you used a different method before, do the following: Use first order integrated rate law to find k: ln(220/116)=-kt ln(220/116)=-k(50) k=1.28 x 10^-2