### CHAPTER 13 REGULAR HOMEWORK

### KINETICS I

**Question 1**

Cyclobutane can decompose to form ethylene: C

_{4}H_{8}(g) → 2C_{2}H_{4}(g). The cyclobutane concentration can be measured as a function of time (shown below) by mass spectrometry . Which of the following is the rate of reaction?Select one or more:

**Explanation**

It is -(rate of change of cylcobutane) because cylcobutane is the reactant, and is being used up. It is .5(rate of change of ethylene) because ethylene is the product (is being created, thus positive value) and has a coefficient of 2 (need to divide by 2 to get rate of reaction).

**Question 2**

Calculate the average rate of reaction between 9.0 and 56.0 s (units mol/L/s).

**Explanation**

To find M values, look at the y-axis value for each time given (9s and 56s)

**Question 3**

Calculate the instantaneous rate of reaction at 63 s (units mol/L/s). (It is a good idea to print the page in order to draw an accurate tangent line.)

**Explanation**

Draw a tangent (only touches curve at one point) at time given (63s in this case) and then find slope of tangent line.

**Question 4**

Calculate the initial rate of reaction (units mol/L/s).

**Explanation**

Find slope of the tangent line at t=0.

**Question 5**

Calculate the instantaneous rate of formation of ethylene at 60.0 s (units mol/L/s).

**Explanation**

Find instantaneous rate of reaction at 60s and multiply by 2 since coefficient of ethylene is 2.

**Question 6**

Under certain conditions, biphenyl, C

_{12}H_{10}, can be produced by the decomposition of cyclohexane, C_{6}H_{12}(see below). The table represents part of the data obtained in kinetics experiments. What is the biphenyl concentration at 1.00 s?**Explanation**

First, find the rate of C6H12 concentration change in 1s:

Then divide by 2 to get rate of reaction per second. This is because the coefficient of C6H12 is 2.

At 1s, the concentration has increased from 0 to 0.0205M.

**Question 7**

What is the cyclohexane concentration when the biphenyl concentration is 0.051 M?

**Explanation**

**Question 8**

What is the hydrogen concentration when the biphenyl concentration is 0.068 M?

**Explanation**

This one’s easy! Since there are 7 moles of H2 for every mole of C12H10 (see coefficients in reaction equation), you multiply the C12H10 concentration by 7 to get H2 concentration.

**Question 9**

Estimate the instantaneous rate of reaction at 0.5 s, by calculating the average rate between 0.00 and 1.00 s?

**Explanation**

Remember the rate of reaction only = -(rate of change of reactant) if the coefficient is one.

(Same work as for Question 6)

**Question 10**

Use the initial rate data in the next question to determine the rate law and rate constant for the gas-phase reaction of nitrogen monoxide and chlorine: 2NO(g) + Cl

_{2}(g) –> 2NOCl(g) Rate law: reaction rate = k[NO]^{x}[Cl_{2}]^{y}What are x and y.Select one or more:

**Explanation**

This Question is very similar to ones in Chapter 13 Basic, remember? 😉

Wordy explanation: •In Experiments 1 and 3 we notice that Cl2 remains constant while NO doubles. Since rate of rxn also doubles, the order of NO (x) is 1. •In Experiments 2 and 4, NO remains constant while Cl2 and rate of rxn double. This means that the order of Cl2 (y) is also 1.

Mathy explanation (yuck!):

**Question 11**

Given:

What is K?

**Explanation**

Solve using rate equation. X and Y have already been determined in Question 10, now just pick an experiment to use for initial rate and [NO] and [Cl2] (we chose Experiment 1).

**Question 12**

For the decomposition of reactant M, determine the order of the reaction from the concentration-time dependence given in the next question. Write 0 for 0 order, 1 for first order, etc. Very useful help for this and the next few problems is found inCh 13 Supplemental Materials (plotting for kinetics) (located below the Ch 13 Supplements).

**Explanation**

We can determine the order by plotting the concentration-time data in excel or on a graphing calculator:

Above we see that plotting this reaction on a [Reactant] v. Time graph will give us a straight line, making it a zero order reaction. For review of plotting and reaction order relationships, go here.

**Question 13**

Determine the rate constant from the concentration-time dependence:

Time (s) | [Reactant](M) |

0.0 | 0.0451 |

2.0 | 0.0421 |

5.0 | 0.0376 |

9.0 | 0.0316 |

15 | 0.0226 |

Use SI units, that is, seconds for time (not included in answer).

**Explanation**

There are two ways to find k: 1. Look at the slope on the graph above (-0.0015). That is equivalent to -k. 2. Use integrated rate law equation for zero order: pick 2 [Reactant] and Time data points and plug n chug! Choo-Choo!

**Question 14**

For the decomposition of reactant M, determine the order of the reaction from the concentration-time dependence given in the next question.

**Explanation**

Just like Q12, plot points in Excel/graphing calculator. This time, we do not get a straight line for our [Reactant] v. Time graph. Bummer! Next, we try plotting ln([Reactant]) v. Time to see if we get a straight line:

Got it! This means that our reaction is 1st order. If you’re not sure why, review here.

**Question 15**

Determine the rate constant from the concentration-time dependence:

Time (s) | [Reactant](M) |

0.0 | 0.0350 |

10 | 0.0223 |

20 | 0.0142 |

50 | 0.0037 |

70 | 0.0015 |

Use SI units, that is, seconds for time (not included in answer).

**Explanation**

There are two ways to find k: 1. Look at the slope on the graph above (-0.045). That is equivalent to -k. 2. Use integrated rate law equation for first order: pick 2 [Reactant] and Time data points and plug n chug! Choo-Choo!

**Question 16**

Determine the order of the reaction from the concentration-time dependence given in the next question. (If needed, a hint is given in the general feedback.)

**Explanation**

Hint, (1) print the graph and draw a smooth curve through the data. (2) Construct a table of time and concentration. (3) You are now at the starting point for a procedure you have already used.

(1) Draw curve:

(2) Obtain points from drawn curve:

Time (s) | Concentration (M) |
---|---|

0 | 0.5 |

20 | 0.3 |

40 | 0.2 |

60 | 0.16 |

80 | 0.125 |

(3) Plot in excel/graphing calculator and determine order as in Q12 and Q14. In this case we get a straight line when plotting 1/[A] v. Time, which tells us that it is a 2nd order rxn. If you’re not sure why this is, review here.

**Question 17**

Determine the rate constant from the concentration-time dependence.

**Explanation**

Use points from the smooth curve that we drew in Q16 and the second order integrated rate law to find k. 🙂

**Question 18**

Calculate the half-life (in s) of a first-order reaction if the concentration of the reactant is 0.0722 M 24.4 s after the reaction starts and is 0.0338 M 52.7 s after the reaction starts.

**Explanation**

(1) Use first order integrated rate law to find k (2) Use the half-life equation for first-order reactions

**Question 19**

Suppose the half-life is 48.6 s for a first order reaction and the reactant concentration is 0.0570 M 61.7 s after the reaction starts. How many seconds after the start of the reaction does it take for the reactant concentration to decrease to 0.0156 M?