Problem 94 - Chemistry The Molecular Science

Question

For the reaction of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) at \(660 \mathrm{~K}\), \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\) \begin{tabular}{lcc} \hline \multicolumn{2}{c} { Concentration (mol/L) } & \\ \hline [NO] & {\(\left[\mathrm{O}_{2}\right]\)} & Rate of Disappearance of \(\mathrm{NO}\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 0.010 & 0.010 & \(2.5 \times 10^{-5}\) \\ 0.020 & 0.010 & \(1.0 \times 10^{-4}\) \\ 0.010 & 0.020 & \(5.0 \times 10^{-5}\) \\ \hline \end{tabular} (a) Determine the order with respect to each reactant. (b) Write the rate equation for the reaction. (c) Calculate the rate constant. (d) Calculate the rate when [NO] \(=0.025 \mathrm{~mol} / \mathrm{L}\) and \(\left[\mathrm{O}_{2}\right]=0.050 \mathrm{~mol} / \mathrm{L}\) (e) If \(\mathrm{O}_{2}\) disappears at a rate of \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}, \mathrm{cal}-\) culate the rate at which NO is consumed. Calculate the rate at which \(\mathrm{NO}_{2}\) is formed.

Step-by-Step Solution

Answer

(a) NO second order, O2 first order; (b) Rate = k[NO]^2[O2]; (c) k = 2.5 × 10^3 mol⁻²L²s⁻¹; (d) Rate = 7.8125 × 10⁻⁴ mol/L/s; (e) Rate of NO = 2.0 × 10⁻⁴ mol/L/s, Rate of NO2 = 2.0 × 10⁻⁴ mol/L/s.

Step 1: Determine Reaction Order for NO

Look at the given data, for cases where \[ \text{[O}_2\text{]} = 0.010 \, \text{mol/L}\], the concentration of NO doubles from 0.010 to 0.020 mol/L. The rate of disappearance of NO increases from \[ 2.5 \times 10^{-5} \, \text{mol/L/s}\] to \[ 1.0 \times 10^{-4} \, \text{mol/L/s}\]. This is a fourfold increase. Thus, the reaction is second order with respect to NO.

Step 2: Determine Reaction Order for O2

Now examine the cases where \[ \text{[NO]} = 0.010 \, \text{mol/L}\]. The concentration of \[ \text{O}_2 \] doubles from 0.010 to 0.020 mol/L, and the rate doubles from \[ 2.5 \times 10^{-5} \, \text{mol/L/s}\] to \[ 5.0 \times 10^{-5} \, \text{mol/L/s}\]. This indicates a first order with respect to \[ \text{O}_2 \].

Step 3: Write the Rate Equation

Based on the determined orders, the rate equation is \[ \text{Rate} = k [\text{NO}]^2 [\text{O}_2]^1\] where \ k \ is the rate constant.

Step 4: Calculate the Rate Constant

Using the rate equation and data from any row, e.g., \\[ \text{[NO]} = 0.010 \, \text{mol/L}, \; \text{[O}_2\text{]} = 0.010 \, \text{mol/L} \; \text{and Rate} = 2.5 \times 10^{-5} \, \text{mol/L/s} \], \solve for \ k: \\[ 2.5 \times 10^{-5} = k (0.010)^2 (0.010) \Rightarrow k = 2.5 \times 10^3 \, \text{mol}^{-2} \text{L}^2 \text{s}^{-1} \].

Step 5: Calculate Rate for Different Concentrations

Substitute \[ \text{[NO]} = 0.025 \, \text{mol/L} \] and \[ \text{[O}_2\text{]} = 0.050 \, \text{mol/L} \] into rate equation: \\[ \text{Rate} = 2.5 \times 10^3 \times (0.025)^2 \times (0.050) \] \Evaluate to determine the rate.

Step 6: Calculate Rate of NO Consumption and NO2 Formation

Given that \[ \text{Rate of disappearance of}\, \text{O}_2 = 1.0 \times 10^{-4} \, \text{mol/L/s} \, \] use stoichiometry to find other rates. For \text{O}_2 disappearance, \[ 2 \] moles of NO are consumed per \[ 1 \] mole of \text{O}_2, so \[ \text{Rate of NO consumption} = 2 \times 1.0 \times 10^{-4} = 2.0 \times 10^{-4} \, \text{mol/L/s} \,\] and the formation of \[ NO_2 \] will match the disappearance of NO.

Key Concepts

Reaction Order

In chemical kinetics, the reaction order indicates how the rate of a reaction depends on the concentration of reactants. For the reaction between nitric oxide (NO) and oxygen (\(\mathrm{O}_{2}\)), the order with respect to each reactant can be determined by observing how changes in concentration affect the rate of reaction.
To find the reaction order for NO, examine cases where the concentration of \([\mathrm{O}_{2}] = 0.010\, \text{mol/L}\). When \([\mathrm{NO}]\) doubles from 0.010 to 0.020 \( \text{mol/L} \), the rate quadruples. This fourfold increase suggests that the reaction is second-order with respect to NO.
For \(\mathrm{O}_{2}\), keep \([\mathrm{NO}] = 0.010\, \text{mol/L}\) and double \([\mathrm{O}_{2}]\) from 0.010 to 0.020 \(\text{mol/L}\). The rate doubles, indicating a first-order dependence on \(\mathrm{O}_{2}\). The overall reaction order is the sum of these individual orders.

Rate Law

The rate law expresses the rate of reaction as a function of the concentration of reactants and a rate constant. The form of the rate law is determined once the orders with respect to each reactant are known.
For the reaction \(2\mathrm{NO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2\mathrm{NO}_{2}(\mathrm{g}) \), based on the orders determined, the rate law is:

• Rate = \( k [\mathrm{NO}]^2 [\mathrm{O}_{2}]^1 \)

Here, \( k \) is the rate constant, \( [\mathrm{NO}] \) and \( [\mathrm{O}_{2}] \) are the concentrations of the reactants. Each concentration is raised to the power of its respective reaction order. This equation allows us to predict how the reaction rate will change with varying concentrations of each reactant.

Rate Constant

The rate constant, symbolized as \( k \), is a crucial part of the rate law. It provides the proportionality factor that relates the rate of reaction to the concentrations of reactants raised to their respective orders.

• To find \( k \), substitute known values of concentration and rate into the rate law and solve.

Using the data point where \([\mathrm{NO}] = 0.010\, \text{mol/L}\), \([\mathrm{O}_{2}] = 0.010\, \text{mol/L}\), and the rate is \(2.5 \times 10^{-5}\, \text{mol/L/s}\), we can solve for \( k \):

• \(2.5 \times 10^{-5} = k (0.010)^2 (0.010)\)
• \(k = 2.5 \times 10^3\, \text{mol}^{-2} \text{L}^2 \text{s}^{-1}\)

This value of \( k \) can then be used to predict reaction rates under different conditions by using the rate law equation.

Stoichiometry

Stoichiometry in chemical kinetics relates the quantities of reactants and products by their coefficients in a balanced chemical equation. This concept helps us understand how the consumption and formation rates of substances are interconnected.
For the reaction \(2\mathrm{NO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2\mathrm{NO}_{2}(\mathrm{g}) \), stoichiometry indicates that:

• 2 moles of NO are consumed for every mole of \(\mathrm{O}_{2}\) consumed.
• 2 moles of \(\mathrm{NO}_{2}\) are formed for every 2 moles of NO consumed.

If the disappearance rate of \(\mathrm{O}_{2}\) is specified, say \(1.0 \times 10^{-4}\, \text{mol/L/s}\), the rate of NO consumption will be twice that, or \(2.0 \times 10^{-4}\, \text{mol/L/s}\), due to the 2:1 stoichiometric ratio. The formation rate of \(\mathrm{NO}_{2}\) matches the consumption rate of NO, based on the balanced equation, satisfying the 1:1 formation to consumption ratio for \(\mathrm{NO}_{2}\).