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 \(\left[\mathrm{O}_{2} \mid=0.035 M\right.\), 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 \(4.65 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (b) The value of the rate constant (k) is approximately \(1.23 \times 10^{-2}\). (c) The units for the rate constant are \(M^{-2} \cdot s^{-1}\). (d) Increasing the concentration of NO by a factor of 1.8 will result in an increase in the rate of the reaction to approximately \(3.01 \times 10^{-4} \mathrm{M} / \mathrm{s}\).

Step-by-step solution

(a) Rate of Disappearance of O2

The rate law for this reaction is given by: \[\mathrm{R}[\mathrm{NO}] = k(\mathrm{NO})^2(\mathrm{O}_{2}) \] From the stoichiometry of the given reaction, we can deduce that 1 mole of O2 is consumed for every 2 moles of NO. Therefore, the rate of disappearance for O2 should be half the rate of disappearance for NO: \(\mathrm{R}[\mathrm{O}_{2}] = \frac{1}{2} \mathrm{R}[\mathrm{NO}]\) Now, we can simply plug in the given rate of disappearance of NO: \(\mathrm{R}[\mathrm{O}_{2}] = \frac{1}{2} \times 9.3 \times 10^{-5} = 4.65 \times 10^{-5} \mathrm{M} / \mathrm{s}\) So, the rate of disappearance of O2 at this moment is \(4.65 \times 10^{-5} \mathrm{M} / \mathrm{s}\).

(b) Value of the Rate Constant

We can find the rate constant (k) by rearranging the rate law formula: \(k = \frac{\mathrm{R}[\mathrm{NO}]}{(\mathrm{NO})^2(\mathrm{O}_{2})}\) Now, plug in the given concentrations of NO and O2, and the rate of disappearance of NO: \(k = \frac{9.3 \times 10^{-5}}{(0.040)^2 (0.035)}\) Calculating the value, we get: \(k \approx 1.23 \times 10^{-2}\) Therefore, the value of the rate constant (k) is approximately \(1.23 \times 10^{-2}\).

(c) Units of the Rate Constant

To find the units of the rate constant, we can use the rate law and rearrange it: \(k = \frac{\mathrm{R}[\mathrm{NO}]}{[\mathrm{NO}]^2[\mathrm{O}_{2}]}\) The units for the numerator are M/s. The units for the denominator are the concentration of NO squared, which means M^2 and the concentration of O2, which is M. In total, the units of the denominator are M^3. So, the units for the rate constant will be: \(Units(k) = \frac{M/s}{M^3} = M^{-2} \cdot s^{-1}\) Thus, the units for the rate constant are \(M^{-2} \cdot s^{-1}\).

(d) Effect of Increasing NO Concentration

If the concentration of NO is increased by a factor of 1.8, let's look how it will affect the rate of the reaction: Since the reaction is second order in NO, the rate will increase by the square of the factor by which the concentration increases: \(\text{New rate} = (1.8)^2 \times \text{Initial rate}\) \(\text{New rate} = 3.24 \times \mathrm{R}[\mathrm{NO}]\) Calculating the new rate: \(\text{New rate} = 3.24 \times 9.3 \times 10^{-5} = 3.01 \times 10^{-4} \mathrm{M} / \mathrm{s}\) Therefore, increasing the concentration of NO by a factor of 1.8 will result in an increase in the rate of the reaction to approximately \(3.01 \times 10^{-4} \mathrm{M} / \mathrm{s}\).

Key concepts

Rate Law

The rate law is a mathematical equation that relates the rate of a chemical reaction to the concentration of its reactants. In simple terms, it shows us how the speed of a reaction depends on the amounts of the chemicals we start with. For this specific reaction, the rate law is expressed as:

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

Here, the rate of disappearance of nitric oxide (NO) is influenced by the concentration of NO and oxygen \((O_2)\). The exponents of the concentrations indicate the reaction order for each substance, showing how sensitive the reaction rate is to changes in each reactant's concentration.

Rate Constant

The rate constant, symbolized as \(k\), is a crucial part of the rate law and determines how fast a reaction proceeds at a given temperature. It is a constant factor specific to a particular reaction and conditions. For our reaction,

• We calculated \(k\) using:\[k = \frac{9.3 \times 10^{-5}}{(0.040)^2 (0.035)}\]
• The value came out to be approximately \(1.23 \times 10^{-2}\).

The units of the rate constant provide insights into the overall order of the reaction. In this case, the units were determined to be \(M^{-2} \cdot s^{-1}\), indicating a third-order reaction across all reactants combined.

Order of Reaction

The order of a reaction tells us how the rate changes when we change the concentration of the reactants. It is crucial as it depicts the effect of reactant concentration on the speed of the reaction. In this problem, the reaction is second order in nitric oxide (NO) and first order in oxygen \((O_2)\). This translates to:

• For NO, the rate is proportional to the square of its concentration.
• For \(O_2\), the rate is proportional to its concentration raised to the power of one.

The overall reaction order is the sum of these individual orders, which in this case, adds up to three (2 for NO plus 1 for \(O_2\)). This information helps us predict how changes in concentration affect the reaction rate.

Effect of Concentration on Rate

Concentration changes can significantly impact reaction rates. Knowing the reaction order allows us to calculate these effects precisely. In this example, when the concentration of NO is increased by a factor of 1.8, the reaction rate will increase as well. Specifically, since the reaction is second order in NO, the rate increases by the square of the factor of concentration increase.

• Initial Factor: 1.8, so Rate Change: \[(1.8)^2 = 3.24\]
• New Rate: \[3.24 \times 9.3 \times 10^{-5} = 3.01 \times 10^{-4} \mathrm{M} / \mathrm{s}\]

Hence, increasing the NO concentration by 1.8 times results in the reaction rate increasing approximately to \(3.01 \times 10^{-4} \mathrm{M} / \mathrm{s}\), showcasing the strong impact concentration has on reaction kinetics.