Question

Consider the reaction between nitrogen oxide and oxygen. $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ The rate of the reaction is followed by measuring the appearance of \(\mathrm{NO}_{2}\). The following data are obtained: $$ \begin{array}{cccc} \hline \text { Expt. } & \text { [NO] } & {\left[\mathrm{O}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \text { min }) \\ \hline 1 & 0.0244 & 0.0372 & 2.20 \times 10^{-3} \\ 2 & 0.0244 & 0.0122 & 7.23 \times 10^{-4} \\ 3 & 0.0244 & 0.0262 & 1.55 \times 10^{-3} \\ 4 & 0.0732 & 0.0372 & 1.98 \times 10^{-2} \\ \hline \end{array} $$ (a) What is the order of the reaction with respect to each reactant? (b) Write the rate expression for the reaction. (c) Calculate \(k\) for the reaction. What are the units for \(k\) ? (d) What is the rate of the reaction at the temperature of the experiment when \([\mathrm{NO}]=0.0100 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.0462 \mathrm{M} ?\)

Short answer

Question: Determine the order of the reaction with respect to each reactant, write the rate expression for the reaction, calculate the rate constant (k) and its units, and calculate the rate of the reaction when [NO] = 0.0100 M and [O2] = 0.0462 M. Answer: The order of the reaction with respect to NO is 2 (second-order), and with respect to O2 is 1 (first-order). The rate expression for the reaction is: Rate = k[NO]^2[O2]^1. The rate constant (k) is 1.23 × 10^2 (mol / L)^{-1} min^{-1}. The rate of the reaction when [NO] = 0.0100 M and [O2] = 0.0462 M is 5.68 × 10^{-4} mol / L * min.

Step-by-step solution

(a) Determine the order of the reaction with respect to each reactant

To determine the order of the reaction for each reactant, we need to look at how the initial rate of reaction changes when concentrations of reactants are varied. We will use the initial rate method, where we can compare different experiments to see the relationship between the rate and the concentration: 1. Comparing experiments 1 and 2, we can see that the concentration of NO remains constant while the concentration of O2 is changed. This will allow us to determine the order of the reaction with respect to O2. Initial rate of Expt.1 / Initial rate of Expt.2 = (2.20 × 10^{-3}) / (7.23 × 10^{-4}) = 3.043 Since [O2] in Expt.1 / [O2] in Expt.2 = 0.0372 / 0.0122 = 3.049 which is nearly equal to the ratio of the initial rates, we can conclude that the order of the reaction with respect to O2 is 1 (first-order). 2. Comparing experiments 1 and 4, we can see that the concentration of O2 remains constant while the concentration of NO is changed. This will allow us to determine the order of the reaction with respect to NO. Initial rate of Expt.4 / Initial rate of Expt.1 = (1.98 × 10^{-2}) / (2.20 × 10^{-3}) = 9.000 Since [NO] in Expt.4 / [NO] in Expt.1 = 0.0732 / 0.0244 = 3.000 and the square of this value (3.000^2) equals 9, we can conclude that the order of the reaction with respect to NO is 2 (second-order).

(b) Write the rate expression for the reaction

Now that we have determined the orders of the reaction with respect to NO and O2, we can write the rate expression. The general form of a rate expression is: Rate = k[NO]^m[O2]^n Where m and n are the orders of the reaction with respect to NO and O2, and k is the rate constant. From part (a), we found that the order of the reaction with respect to NO is 2 and with respect to O2 is 1. Therefore, the rate expression is: Rate = k[NO]^2[O2]^1

(c) Calculate k for the reaction and find its units

To calculate the rate constant k, we will use the data from any of the experiments and the rate expression we determined in part (b). We will choose experiment 1: Rate = k[NO]^2[O2]^1 2.20 × 10^{-3} mol / L * min = k(0.0244)^2(0.0372) Now, solve for k: k = 1.23 × 10^2 (mol / L)^{-1} min^{-1} The units of k are (mol / L)^{-1} min^{-1}.

(d) Calculate the rate of the reaction for given concentrations of NO and O2

Now, we will calculate the rate of the reaction when [NO] = 0.0100 M and [O2] = 0.0462 M using the rate expression and the value of k: Rate = k[NO]^2[O2]^1 Rate = (1.23 × 10^2) (0.0100)^2 (0.0462) Rate = 5.68 × 10^{-4} mol / L * min Thus, the rate of the reaction when [NO] = 0.0100 M and [O2] = 0.0462 M is 5.68 × 10^{-4} mol / L * min.