## Key Concepts and Summary

Rate laws (*differential rate laws*) provide a mathematical description of how changes in the concentration of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the concentration of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.

## Practice Problems: Rate Laws

How do the rate of a reaction and its rate constant differ?

## Solution

A *rate of reaction* describes how quickly a reaction is proceeding – at any given moment – and is measured in units of $\frac{concentration}{time}$. A *rate constant* is __indirectly__ related to the rate of reaction – a larger rate constant means the reaction will likely be faster – but the actual rate will also depend on the instantaneous concentrations of the reactants. Rate constants have units that are based on concentration and time, but are determined by the overall order of the reaction.

Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:

(a) What is the order of the reaction with respect to that reactant?

## Solution

2

(b) Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?

## Solution

1

Tripling the concentration of a reactant increases the rate of a reaction nine-fold. With this knowledge, answer the following questions:

(a) What is the order of the reaction with respect to that reactant?

## Solution

2

(b) Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four-fold. What is the order of the reaction with respect to that reactant?

## Solution

1

How will the rate of reaction change for the process: $CO(g)+NO_2(g)⟶CO_2(g)+NO(g)$ if the rate law for the reaction is $\text{rate}=k[NO_2]^2$?

(a) Decreasing the pressure of NO_{2} from 0.50 atm to 0.250 atm.

## Solution

The process reduces the rate by a factor of 4.

(b) Increasing the concentration of CO from 0.01 *M* to 0.03 *M*.

## Solution

Since CO does not appear in the rate law (0^{th} order), the rate is not affected.

How will each of the following affect the rate of the reaction: $CO(g)+NO_2(g)⟶CO_2(g)+NO(g)$ if the rate law for the reaction is $\text{rate}=k[NO_2][CO]$?

(a) Increasing the pressure of NO_{2} from 0.1 atm to 0.3 atm

## Solution

Since the reaction is first-order in NO_{2}, tripling( x3) the concentration will also triple (x3^{1}) the rate.

(b) Increasing the concentration of CO from 0.02 *M* to 0.06 *M*.

## Solution

Since the reaction is first-order in CO, tripling( x3) the concentration will also triple (x3^{1}) the rate.

Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction $NO+O_3⟶NO_2+O_2$ is first order with respect to both NO and O_{3} with a rate constant of $2.20×10^7$ L/mol/s. What is the instantaneous rate of disappearance of NO when [NO] = $3.3×10^{-6}\;M$ and [O_{3}] = $5.9×10^{-7}\;M$?

## Solution

$4.3×10^{-5}$ mol/L/s

Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:

What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 *M*?

## Solution

We can use the provided rate law and concentration to solve for the rate of reaction: $$\text{rate}=4.85\times 10^{-2}\;\text{day}^{-1}\;[^{32}P] \\ \text{rate}=4.85\times 10^{-2}\;\text{day}^{-1}\;[0.0033\:M] \\ \text{rate}=1.60\times 10^{-4} \frac{M}{day}$$ Since the rate of production of electrons is 1:1 with the overall rate, the value is the same.

The rate constant for the radioactive decay of ^{14}C is $1.21×10^{-4}$ year^{−1}. The products of the decay are nitrogen atoms and electrons (beta particles):

What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of $6.5×10^{-9}\;M$?

## Solution

We can use the provided rate law and ^{14}C content to determine the rate:
$$\text{rate}=k[^{14}_6C] \\
\text{rate}=(1.21 \times 10^{-4}\: year^{-1})[6.5 \times 10^{-9} M] \\
\text{rate} = 7.9\times 10^{-13} \frac{M}{year}$$

The decomposition of acetaldehyde is a second order reaction with a rate constant of $4.71×10^{-8}$ L mol^{−1} s^{−1}. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of $5.55×10^{-4}\;M$?

## Solution

First, we need to write the rate law, knowing that it is a second-order reaction. We can infer that (as a decomposition) the only reactant is the acetaldedyde: $$ \text{rate} = k[acetaldehyde]^2$$ Then, we can substitute in the values given and solve for the rate of reaction: $$ \text{rate} = \left(4.71×10^{-8}\: \frac{L}{mol \cdot s}\right)[5.55×10^{-4}\;M]^2 \\ \text{rate} = 1.45 \times 10^{-14}\; \frac{M}{s}$$

Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men:

Run 1 | Run 2 | Run 3 | |

[C_{2}H_{5}OH] (M) |
$4.4×10^{-2}$ | $3.3×10^{-2}$ | $2.2×10^{-2}$ |

Rate (mol L^{−1} h^{−1}) |
$2.0×10^{-2}$ | $2.0×10^{-2}$ | $2.0×10^{-2}$ |

Determine the rate law, the rate constant, and the overall order for this reaction.

## Solution

From the data provided, the initial concentration of ethanol has no effect on the rate of reaction. Therefore the reaction is **zero-order**: $\text{rate} = k$. Using the data from any experiment in the table, $\text{rate} = k = 2.0\times 10^{-2} \frac{M}{h}$. Therefore, the complete rate law is:
$$\text{rate} = 2.0\times 10^{-2} \frac{M}{h}$$

Under certain conditions the decomposition of ammonia on a metal surface gives the following data:

Run 1 | Run 2 | Run 3 | |

[NH_{3}] (M) |
$1.0×10^{-3}$ | $2.0×10^{-3}$ | $3.0×10^{-3}$ |

Rate (mol L^{−1} h^{−1}) |
$1.5×10^{-6}$ | $1.5×10^{-6}$ | $1.5×10^{-6}$ |

Determine the rate law, the rate constant, and the overall order for this reaction.

## Solution

From the data provided, the initial concentration of ammonia has no effect on the rate of reaction. Therefore the reaction is **zero-order**: $\text{rate} = k$. Using the data from any experiment in the table, $\text{rate} = k = 1.5\times 10^{-6} \frac{M}{h}$. Therefore, the complete rate law is:
$$\text{rate} = 1.5\times 10^{-6} \frac{M}{h}$$

Nitrosyl chloride, NOCl, decomposes to NO and Cl_{2}.

Determine the rate law, the rate constant, and the overall order for this reaction from the following data:

Run 1 | Run 2 | Run 3 | |

[NOCl] (M) |
0.10 | 0.20 | 0.30 |

Rate (mol L^{−1} h^{−1}) |
$8.0×10^{-10}$ | $3.2×10^{-9}$ | $7.2×10^{-9}$ |

## Solution

Comparing runs 1 & 2, as the [NOCl] is doubled, the rate has gone up by a factor of 4. 2^{2} = 4, so this reaction is **second order** with respect to [NOCl]. The rate law is therefore:
$$\text{rate} = k[NOCl]^2$$
Substituting in the values from any run and solving for *k* gives *k* = $8.0×10^{-8} \frac{1}{M\cdot h}$, and the complete rate law is:
$$\text{rate} = \left( 8.0×10^{-8} \frac{1}{M\cdot h} \right)[NOCl]^2$$

From the following data, determine the rate law, the rate constant, and the order with respect to $A$ for the reaction $A⟶2C$.

Run 1 | Run 2 | Run 3 | |

[A] (M) |
$1.33×10^{-2}$ | $2.66×10^{-2}$ | $3.99×10^{-2}$ |

Rate (mol L^{−1} h^{−1}) |
$3.80×10^{-7}$ | $1.52×10^{-6}$ | $3.42×10^{-6}$ |

## Solution

Comparing Runs 1 & 2, the concentration of A doubled, the rate has gone up by a factor of 4. 2^{2} = 4, so this reaction is **second order** with respect to A. The rate law is therefore:
$$\text{rate} = k[A]^2$$
Substituting in the values from any run and solving for *k* gives *k* = $2.15 \times 10^{-3} \frac{1}{M \cdot h}$, and the complete rate law is:
$$\text{rate} = \left( 2.15 \times 10^{-3} \frac{1}{M \cdot h} \right)[A]^2$$

Nitrogen monoxide reacts with chlorine according to the equation:

The following initial rates of reaction have been observed for certain reactant concentrations:

[NO] (mol/L^{1}) |
[Cl_{2}] (mol/L) |
Rate (mol L^{−1} h^{−1}) |
---|---|---|

0.50 | 0.50 | 1.14 |

1.00 | 0.50 | 4.56 |

1.00 | 1.00 | 9.12 |

What is the rate law that describes the rate’s dependence on the concentrations of NO and Cl_{2}? What is the rate constant? What are the orders with respect to each reactant?

## Solution

Comparing Rows 1 & 2 (where [Cl_{2}] is constant, we see that as [NO] increases by x2, the rate increases by x4. This means the reaction is **second order in [NO]**.

Comparing Rows 2 & 3 (where [NO] is constant, we see that as [Cl_{2}] increases by x2, the rate increases by x2 as well. Therefore, the reaction is **first order in [Cl _{2}]**.

The rate law is therefore: $$\text{rate}=k[NO]^2 [Cl_2]$$ By substituting in data from any run, we can find $k = 9.1 \frac{1}{M^2 h}$, making the complete rate law: $$\text{rate}=\left( 9.1 \frac{1}{M^2 h} \right)[NO]^2 [Cl_2]$$

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation:

$$H_2(g)+2NO(g)⟶N_2O(g)+H_2O(g)$$Determine the rate law, the rate constant, and the orders with respect to each reactant from the following data:

Run 1 | Run 2 | Run 3 | |

[NO] (M) |
0.30 | 0.60 | 0.60 |

[H_{2}] (M) |
0.35 | 0.35 | 0.70 |

Rate (mol L^{−1} s^{−1}) |
$2.835×10^{-3}$ | $1.134×10^{-2}$ | $2.268×10^{-2}$ |

## Solution

Comparing Runs 1 & 2 (H_{2} is constant), as [NO] doubles, rate increases by x4: therefore the reaction is **second order in NO**. Comparing Runs 2 & 3 ([NO] is constant), as [H_{2}] doubles, the rate doubles, therefore the reaction is ** first order in H _{2}**. The rate law is:
$$\text{rate}=k[NO]^2 [H_2 ]$$
Substituting data from any run into this gives $ k = 9.0 \times 10^{-2}\: \frac{1}{M^2 \cdot s}$. The complete rate law is:
$$\text{rate}=\left( 9.0 \times 10^{-2}\: \frac{1}{M^2 \cdot s} \right)[NO]^2 [H_2 ]$$

For the reaction $A⟶B+C$, the following data were obtained at 30 °C:

Run 1 | Run 2 | Run 3 | |

[A] (M) |
0.230 | 0.356 | 0.557 |

Rate (mol L^{−1} s^{−1}) |
$4.17×10^{-4}$ | $9.99×10^{-4}$ | $2.44×10^{-3}$ |

(a) What is the order of the reaction with respect to [*A*], and what is the rate law?

## Solution

The rate law is second order in A and is written as $\text{rate}=k[A]^2 $. After solving part (b), the complete rate law is $\text{rate}=\left( 7.88×10^{-3} \frac{1}{M\cdot s} \right)[A]^2 $

(b) What is the rate constant?

## Solution

$k = 7.88×10^{-3} \frac{1}{M\cdot s}$

For the reaction $Q⟶W+X$, the following data were obtained at 30 °C:

Run 1 | Run 2 | Run 3 | |

[Q]_{initial} (M) |
0.170 | 0.212 | 0.357 |

Rate (mol L^{−1} s^{−1}) |
$6.68×10^{-3}$ | $1.04×10^{-2}$ | $2.94×10^{-2}$ |

(a) What is the order of the reaction with respect to [*Q*], and what is the rate law?

## Solution

The concentrations in this question are not even multiples (not an exact x2, x3 etc) so it is easier to solve this using the ratio method (Here, using Runs 1 & 2): $$ \left( \frac{0.212}{0.170} \right)^n = \frac{1.04 \times 10^{-2}}{6.68 \times 10^{-3}} \\ \left( 1.24_{70588} \right)^n = 1.55_{6886} \\ ln\left( (1.24_{70588})^n \right) = ln (1.55_{6886}) \\ n \cdot ln(1.24_{70588})= ln (1.55_{6886}) \\ n = \frac{ln (1.55_{6886})}{ln(1.24_{70588})} \\ n = 2.00_{5037} \\ n = 2 $$ The rate law is second order in Q and is written as $\text{rate}=k[Q]^2 $. After solving part (b), the complete rate law is $\text{rate}=\left( 0.231 \frac{1}{M\cdot s} \right)[Q]^2 $

(b) What is the rate constant?

## Solution

$k = 0.231 \frac{1}{M\cdot s}$

The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N_{2}O_{5}, dissolved in chloroform, CHCl_{3}, is $6.2×10^{-4}$ min^{−1}.

What is the rate of the reaction when [N_{2}O_{5}] = 0.40 *M*?

## Solution

(a) $2.5×10^{-4}$ mol/L/min

The annual production of HNO_{3} in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.

(a)$4NH_3(g)+5O_2(g)⟶4NO(g)+6H_2O(g)$

(b)$2NO(g)+O_2(g)⟶2NO_2(g)$

(c)$3NO_2(g)+H_2O(l)⟶2HNO_3(aq)+NO(g)$

The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O_{2}, what is the rate of formation of NO_{2} when the oxygen concentration is 0.50 *M* and the nitric oxide concentration is 0.75 *M*? The rate constant for the reaction is $5.8×10^{-6}$ L^{2} mol^{−2} s^{−1}.

## Solution

Since the set of reactions given is a stepwise mechanism, the rate law (for each step at least) can be determined from the stoichiometry of the elementary step.
For Reaction (b) then, $ rate = k[NO]^2 [O_2 ]$.

Substituting in the values given in the question:
$$ rate = \left( 5.8 \times 10^{-6}\: \frac{M^2}{s} \right) [0.75 M]^2 [0.50 M] \\
rate = 1.6_{31} \times 10^{-6} M/s \\
rate = 1.6 \times 10^{-6} M/s $$

The following data have been determined for the reaction:

Run 1 | Run 2 | Run 3 | |

$[I^-]_{initial}\;(M)$ | 0.10 | 0.20 | 0.30 |

$[OCl^-]_{initial}\;(M)$ | 0.050 | 0.050 | 0.010 |

Rate (mol L^{−1} s^{−1}) |
$3.05×10^{-4}$ | $6.20×10^{-4}$ | $1.83×10^{-4}$ |

Determine the rate law and the rate constant for this reaction.

## Solution

Comparing data from Runs 1 and 2:

$[OCl^- ]$ is constant. $[I^-]$ doubles, and the rate also doubles (approximately). So $ 2^m=2$ and *m* = 1. The reaction is first order in iodide.

Comparing data from Runs 1 and 3:

Both concentrations change (there is no pair where $[I^-]$ is constant) but we can solve it with the ratio method and incorporating the order determined for iodide above – using Runs 1 and 3:
$$ \frac{rate\: A}{rate\: B} = \frac{[I^- ]^1 _A [OCl^-]^m _A}{[I^- ]^1 _B [OCl^-]^m _B} \\
\frac{3.05 \times 10 {-4}}{1.83 \times 10^{-4}} = \frac{[0.10]^1 [0.050]^m}{[0.30]^1 [0.010]^m} \\
1.66_{6667} = 0.33_{3333} \left( \frac{0.050}{0.010} \right)^m \\
log(\frac{1.66_{6667}}{0.33_{3333}}) = m \cdot log(\frac{0.050}{0.010}) \\
m = 1 $$
Therefore the rate constant has the form $ rate = [I^- ] [OCl^- ] $.

The rate constant can be found by using the rate and concentration data from any run: $k = 6.1 \times 10^{-2} \frac{1}{M\cdot s} $.

The full rate law for this reaction is: $rate = \left( 6.1 \times 10^{-2} \frac{1}{M\cdot s} \right) [I^- ] [OCl^- ] $.