Problem 29 - Chemistry The Central Science In Si Units

Question

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(55^{\circ} \mathrm{C}\) the rate constant is \(4.12 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.050 \mathrm{M} ?(\mathbf{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is tripled to \(0.150 \mathrm{M} ?(\mathbf{d})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is reduced by \(10 \%\) to \(0.045 \mathrm{M} ?\)

Step-by-Step Solution

Answer

Rate law: \(\text{rate} = k [\mathrm{N}_{2} \mathrm{O}_{5}]\). Rate is \(2.06 \times 10^{-4}\, \mathrm{M/s}\) at \(0.050 \mathrm{M}\); triples when concentration triples; decreases by 10\% when concentration decreases by 10\%.

Step 1: Identify the Rate Law

For this problem, the rate law is based on the order of the reaction with respect to the reactant, \(\mathrm{N}_{2} \mathrm{O}_{5}\). Since it is first order, the rate law can be written as \( ext{rate} = k [\mathrm{N}_{2} \mathrm{O}_{5}]\), where \(k\) is the rate constant.

Step 2: Calculate the Rate of Reaction for Part (b)

Using the rate law \( ext{rate} = k [\mathrm{N}_{2} \mathrm{O}_{5}]\), substitute \(k = 4.12 \times 10^{-3}\, \mathrm{s}^{-1}\) and \( [\mathrm{N}_{2} \mathrm{O}_{5}] = 0.050 \, \mathrm{M}\) to find the rate of reaction. \[\text{rate} = (4.12 \times 10^{-3}\, \mathrm{s}^{-1}) (0.050 \, \mathrm{M}) = 2.06 \times 10^{-4}\, \mathrm{M/s}.\]

Step 3: Calculate the New Rate for Part (c)

Since the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is tripled, the new concentration is \(0.150\, \mathrm{M}\). Substitute this into the rate law:\[\text{new rate} = (4.12 \times 10^{-3}\, \mathrm{s}^{-1}) (0.150 \, \mathrm{M}) = 6.18 \times 10^{-4}\, \mathrm{M/s}.\]The rate triples because the reaction is first-order and the concentration is tripled.

Step 4: Calculate the New Rate for Part (d)

If the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is reduced by 10\%, then the new concentration becomes \(0.045 \, \mathrm{M}\). Substitute this new concentration into the rate law:\[\text{new rate} = (4.12 \times 10^{-3}\, \mathrm{s}^{-1}) (0.045 \, \mathrm{M}) = 1.854 \times 10^{-4}\, \mathrm{M/s}.\]The rate decreases proportionally by 10\% since the reaction is first-order.

Key Concepts

Understanding Rate Law

Chemical reactions often occur at different speeds, or rates. To understand how fast a reaction is happening, chemists use what is called a **rate law**. A rate law expresses the rate of a reaction in terms of the concentration of the reactants and a constant called the rate constant. In essence, the rate law tells us how the concentration of one or more reactants affects the speed of a chemical reaction.

For instance, in a reaction where the rate depends solely on the concentration of one reactant \(A\), the rate law can be written as: \[\text{rate} = k [A]^n\] Here, \(k\) is the rate constant, and the exponent \(n\) represents the order of the reaction with respect to that reactant.

In the provided exercise, the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). This simplifies the rate law to \[\text{rate} = k [\mathrm{N}_{2} \mathrm{O}_{5}]\] indicating that the rate is directly proportional to the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\).

What Defines a First Order Reaction

A **first order reaction** is a chemical reaction where the rate depends linearly on the concentration of only one reactant. This means that if you were to double the concentration of that reactant, the rate of reaction would also double. Similarly, reducing the concentration by half would cut the rate of reaction by half.

• First-order reactions have a rate law of the form: \[\text{rate} = k [\text{Reactant}]\]
• The unit of the rate constant \(k\) in first order reactions is \(s^{-1}\), which reflects that the rate is a change in concentration per unit time.

This type of reaction is often used to express processes such as radioactive decay and many simple decomposition reactions. In our exercise about \(\mathrm{N}_{2} \mathrm{O}_{5}\), the observation that tripling the concentration also triples the rate confirms that the reaction is first order. It's straightforward but important: the speed of the reaction is tied directly to the concentration of just one reactant.

Deciphering the Rate Constant

The **rate constant** \(k\) is a significant part of the rate law. It is a proportionality constant that links the rate of reaction to the concentrations of the reactants. Each reaction at a given temperature has a unique rate constant.

• For first order reactions, the units of \(k\) are \(s^{-1}\).
• The value of \(k\) gives insight into the speed of the reaction: a larger \(k\) indicates a faster reaction.

It's crucial to understand that \(k\) is affected by temperature. If you increase the temperature, the rate constant typically increases, speeding up the reaction. In the exercise, the reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) has a rate constant of \(4.12 \times 10^{-3} \, \mathrm{s}^{-1}\) at 55°C, highlighting the specific condition under which the reaction rate was measured and showing that reactions can be highly temperature-dependent.