Question

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\) . When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M},\) the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\) . (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8\(?\)

Short answer

(a) The rate of disappearance of O2 at this moment is \(4.65 \times 10^{-5} M/s\). (b) The value of the rate constant k is 1.66 \(M^{-1}s^{-1}\). (c) The units of the rate constant are \(M^{-1}s^{-1}\). (d) If the concentration of NO is increased by a factor of 1.8, the rate of the reaction will increase by a factor of 3.24.

Step-by-step solution

Write the rate law equation

The reaction is second order in NO and first order in O2. The rate law equation can be written as: Rate = k[NO]^2[O2]

Calculate the rate of disappearance of O2

Since we are given the rate of disappearance of NO, we can use the stoichiometric relationship to find the rate of disappearance of O2. Observe that for every 2 moles of NO, 1 mole of O2 is consumed. We can now write: \(Rate_{O_2} = \frac{1}{2} \times Rate_{NO}\) Now, substitute the given rate of disappearance of NO and find the rate of disappearance of O2: \(Rate_{O_2} = \frac{1}{2} \times 9.3 \times 10^{-5} = 4.65 \times 10^{-5} M/s\) Answer (a): The rate of disappearance of O2 at this moment is \(4.65 \times 10^{-5} M/s\).

Find the value of the rate constant

We can use the rate law equation to find the value of the rate constant, k. Substitute the given concentrations and the known rate of disappearance of NO: \(9.3 \times 10^{-5} = k (0.040)^2(0.035)\) Now, solve for k: \(k = \frac{9.3 \times 10^{-5}}{(0.040)^2(0.035)} = 1.66 M^{-1}s^{-1}\) Answer (b): The value of the rate constant k is 1.66 M^{-1}s^{-1}.

Find the units of the rate constant

The rate constant, k, has already been determined with its units in the previous step. Answer (c): The units of the rate constant are M^{-1}s^{-1}.

Determine the effect of increasing the concentration of NO on the rate

We are asked to find what would happen to the rate if the concentration of NO were increased by a factor of 1.8. We can use the rate law equation to find the new rate: \(NewRate = k(NewNO)^2[O_2]\) Since the concentration of NO is increased by a factor of 1.8: \(NewRate = k(1.8[NO])^2 [O_2] = k(3.24[NO]^2) [O_2]\) We know the initial rate was equal to k[NO]^2[O2]. Therefore, the new rate can be expressed as: \(NewRate = 3.24 \times Rate\) Answer (d): If the concentration of NO is increased by a factor of 1.8, the rate of the reaction will increase by a factor of 3.24.

Key concepts

Rate Law

Understanding the rate law is fundamental in chemical kinetics as it expresses how the rate of a reaction depends on the concentration of reactants. For the given reaction between nitrogen monoxide (\text{NO}) and oxygen (\text{O}_2) to produce nitrogen dioxide (\text{NO}_2), the rate law is given by the expression
\[ \text{Rate} = k[\text{NO}]^2[\text{O}_2] \]
where \(k\) is the rate constant, and the exponents 2 and 1 refer to the reaction order with respect to \(\text{NO}\) and \(\text{O}_2\), respectively. This tells us that the reaction is second order in \(\text{NO}\) and first order in \(\text{O}_2\), meaning that the rate of the reaction is directly proportional to the square of the concentration of \(\text{NO}\) and the first power of the concentration of \(\text{O}_2\). This type of detailed understanding helps to predict how changing the concentrations of reactants will affect the rate of the reaction.

Reaction Order

The reaction order indicates the power dependence of the rate on the concentration of each reactant. In the provided problem, the reaction's order with respect to \(\text{NO}\) is 2, and with respect to \(\text{O}_2\) is 1. This implies that if you double the concentration of \(\text{NO}\), the rate of the reaction will increase by a factor of four, because the reaction is second order with respect to \(\text{NO}\). Similarly, if \(\text{O}_2\)'s concentration is doubled, the rate will only double, reflecting the first order dependence. This concept is crucial when considering how reactions speed up or slow down when concentrations change.

Rate Constant

The rate constant, denoted as \(k\), is a proportionality constant in the rate law that is specific to a particular reaction at a given temperature. It relates the rate of reaction to the concentration of the reactants. From the solution, we calculated that \(k = 1.66 \text{M}^{-1}\text{s}^{-1}\) for this reaction under the given conditions. Importantly, the value of the rate constant provides insight into the speed of the reaction; a larger \(k\) signifies a faster reaction. The units of the rate constant also give us important information about the overall order of the reaction, which in this case is third order, as the sum of the exponents is 3 (2 for \(\text{NO}\) plus 1 for \(\text{O}_2\)).

Stoichiometry

Stoichiometry is about the quantitative relationships between reactants and products in a chemical reaction. For the reaction in question, the stoichiometric coefficients in the balanced chemical equation indicate that two moles of \(\text{NO}\) react with one mole of \(\text{O}_2\) to form two moles of \(\text{NO}_2\). The stoichiometry of the reaction allows us to link the rates of disappearance of different reactants. For instance, since \(\text{O}_2\) is consumed at half the rate that \(\text{NO}\) is consumed, we can use the rate of disappearance of \(\text{NO}\) to find that of \(\text{O}_2\), as shown in the provided solution. Additionally, understanding stoichiometry is essential when calculating the theoretical yield of a product or the required quantities of reactants for a reaction.