Problem 5 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

For the complex reaction of \(\mathrm{NO}\) and \(\mathrm{H}_{2}\) \\[ 2 \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\] the rate equation is given by rate of reaction \(\left.=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right] \text { (Section } 9.5\right)\) (a) What are the orders of the reaction with respect to NO and \(\mathrm{H}_{2}\) ? (b) What is the overall order of the reaction? (c) What will happen to the rate of reaction if: (i) \(\left[H_{2}\right]\) is doubled (ii) \(\left[\mathrm{H}_{2}\right]\) is halved; (iii) \([\mathrm{NO}]\) is doubled (iv) \([\mathrm{NO}]\) is increased by a factor of three?

Step-by-Step Solution

Answer

(a) Order w.r.t NO is 2; w.r.t \(\text{H}_2\) is 1. (b) Overall order is 3. (c) (i) Rate doubles, (ii) Rate halves, (iii) Rate quadruples, (iv) Rate increases by factor of 9.

Step 1: Identify Reaction Orders from the Rate Equation

From the given rate equation: \( \text{rate} = k[\text{NO}]^2[\text{H}_2] \), we identify the order with respect to each reactant. The exponent of \([\text{NO}]\) is 2 and the exponent of \([\text{H}_2]\) is 1. Thus, the order of reaction with respect to NO is 2, and the order with respect to \(\text{H}_2\) is 1.

Step 2: Determine Overall Reaction Order

The overall order of a reaction is the sum of the orders with respect to all reactants. Therefore, the overall reaction order is the sum of the orders with respect to NO (2) and \(\text{H}_2\) (1):\[ 2 + 1 = 3 \]. Thus, the overall order of the reaction is 3.

Step 3: Effect of Doubling \(\left[\text{H}_2\right]\) on the Rate

When \(\left[\text{H}_2\right]\) is doubled, the rate equation becomes: \( \text{rate} = k[\text{NO}]^2(2[\text{H}_2]) \). Since the order with respect to \([\text{H}_2]\) is 1, doubling \([\text{H}_2]\) will double the rate of the reaction. Therefore, the rate will increase by a factor of 2.

Step 4: Effect of Halving \(\left[\text{H}_2\right]\) on the Rate

When \(\left[\text{H}_2\right]\) is halved, the rate equation becomes: \( \text{rate} = k[\text{NO}]^2(0.5[\text{H}_2]) \). Since the order with respect to \([\text{H}_2]\) is 1, halving \([\text{H}_2]\) will halve the rate of the reaction. Therefore, the rate will decrease by a factor of 0.5.

Step 5: Effect of Doubling \([\mathrm{NO}]\) on the Rate

When \([\mathrm{NO}]\) is doubled, the rate equation becomes: \( \text{rate} = k(2[\mathrm{NO}])^2[\mathrm{H}_2] \). The rate is proportional to \([\mathrm{NO}]^2\), so doubling \([\mathrm{NO}]\) will increase the rate by \((2)^2 = 4\) times. Therefore, the rate will increase by a factor of 4.

Step 6: Effect of Tripling \([\mathrm{NO}]\) on the Rate

When \([\mathrm{NO}]\) is increased by a factor of three, the rate equation becomes: \( \text{rate} = k(3[\mathrm{NO}])^2[\mathrm{H}_2] \). The rate is proportional to \([\mathrm{NO}]^2\), so tripling \([\mathrm{NO}]\) will increase the rate by \((3)^2 = 9\) times. Therefore, the rate will increase by a factor of 9.

Key Concepts

Reaction Order

In chemical reactions, the *reaction order* refers to how the concentration of reactants affects the rate of reaction. It is derived from the rate equation and shows the power to which the concentration of each reactant is raised. In the reaction \[2 \,\text{NO(g)} + 2 \,\text{H}_2\,\text{(g)} \rightarrow \text{N}_2\,\text{(g)} + 2 \,\text{H}_2\text{O}\,\text{(g)}\],the rate equation is given by \[\text{rate} = k[\text{NO}]^2[\text{H}_2].\]- The reaction order with respect to NO is 2. This means that if the concentration of NO is doubled, the rate of reaction will increase by \((2)^2 = 4\) times.- The reaction order with respect to \(\text{H}_2\) is 1, indicating a direct proportionality to the concentration change. Doubling \(\text{H}_2\) results in doubling the rate.The specific powers (or exponents) in the rate equation indicate how sensitive the rate is to changes in concentration for each reactant.

Rate Equation

The *rate equation*, sometimes called the rate law, is an expression that relates the rate of a chemical reaction to the concentration of the reactants. For our reaction:\[\text{rate} = k[\text{NO}]^2[\text{H}_2],\]- \(k\) is the rate constant, a fixed value for a given reaction at a specific temperature.- \([\text{NO}]^2\) and \([\text{H}_2]\) are the concentrations of NO and \(\text{H}_2\) , respectively.The rate equation provides crucial insights:- Identifies the order with respect to each reactant by indicating the exponent level. - Helps predict how changes in concentration affect the reaction rate.- Customarily, initial concentrations are used to form these equations since they are measurable and provide consistency.Understanding this relationship is pivotal in controlling and optimizing chemical processes.

Rate of Reaction

The *rate of reaction* is a measure of how quickly a reaction occurs. It is typically expressed as the change in concentration of reactants or products per unit time. For this reaction, we described the rate using the following equation:\[\text{rate} = k[\text{NO}]^2[\text{H}_2].\]This rate equation helps in determining:- Magnitude: How fast or slow the reaction proceeds.- Change: It gives insight into how varying the concentrations of NO and \(\text{H}_2\) alters the reaction pace.Since the rate is dependent on both \([\text{NO}]\) and \([\text{H}_2]\) , if any of these changes, the rate of reaction will also change. This is fundamental in designing experiments and in industrial applications where reaction speed is critical.

Effect of Concentration on Rate

The *effect of concentration on rate* highlights how the reactants' concentrations influence the reaction's speed. From the given rate equation:\[\text{rate} = k[\text{NO}]^2[\text{H}_2],\]several scenarios help illustrate this:- Doubling \([\text{H}_2]\) will double the reaction rate as the order with respect to \(\text{H}_2\) is 1.- If \([\text{H}_2]\) is halved, the rate also halves, indicating a linear relationship.- Doubling \([\text{NO}]\) causes a fourfold increase in rate because it's squared in its term. - Tripling \([\text{NO}]\) results in a ninefold increase, showing the exponential effect of reactant concentration on rate.This sensitivity to concentration is crucial for understanding and controlling chemical reactions in both laboratory and industrial settings.