Question

Gaseous ammonia is made by the reaction $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})$$ Use the information on the formation of \(\mathrm{NH}_{3}\) given in the table to answer the questions that follow. $$\begin{array}{lll}\hline\left[\mathrm{N}_{2}\right](\mathrm{M}) & {\left[\mathrm{H}_{2}\right](\mathrm{M})} & \text { Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\\\\hline 0.030 & 0.010 & 4.21 \times 10^{-5} \\\0.060 & 0.010 & 1.68 \times 10^{-4} \\\0.030 & 0.020 & 3.37 \times 10^{-4} \\\\\hline\end{array}$$ (a) Determine \(n\) and \(m\) in the rate equation: Rate \(=\) \(k\left[\mathrm{N}_{2}\right]^{n}\left[\mathrm{H}_{2}\right]^{m}\) (b) Calculate the value of the rate constant. (c) What is the order of the reaction with respect to \(\left[\mathrm{H}_{2}\right] ?\) (d) What is the overall order of the reaction?

Short answer

(a) \(n=2, m=3\); (b) \(k=4.68 \times 10^3 \, \mathrm{L^4/mol^4/min}\); (c) 3; (d) 5.

Step-by-step solution

Find the Order with Respect to N2

To find the order with respect to \( [\mathrm{N}_2] \), compare experiments where only \( [\mathrm{N}_2] \) changes. Using experiments 1 and 2: \( \frac{0.060}{0.030} = 2 \) doubling \( [\mathrm{N}_2] \) results in \( \frac{1.68 \times 10^{-4}}{4.21 \times 10^{-5}} = 4 \). This infers that \( n = 2 \).

Find the Order with Respect to H2

To find the order with respect to \( [\mathrm{H}_2] \), compare experiments where only \( [\mathrm{H}_2] \) changes. Using experiments 1 and 3: \( \frac{0.020}{0.010} = 2 \) doubling \( [\mathrm{H}_2] \) results in \( \frac{3.37 \times 10^{-4}}{4.21 \times 10^{-5}} = 8 \). This indicates that \( m = 3 \).

Write the Rate Equation

With \( n = 2 \) and \( m = 3 \), the rate equation is: Rate = \( k[\mathrm{N}_2]^2[\mathrm{H}_2]^3 \).

Calculate the Rate Constant k

Using the known concentrations and rate from any experiment, substitute to solve for \( k \). Using experiment 1: \( 4.21 \times 10^{-5} = k(0.030)^2(0.010)^3 \). Thus, \( k = \frac{4.21 \times 10^{-5}}{0.000000009} = 4.68 \times 10^3 \, \mathrm{L^4/mol^4/min} \).

Determine the Order with Respect to \([\mathrm{H}_2]\)

From step 2, the order with respect to \( [H_2] \) is \( m = 3 \).

Determine the Overall Order of the Reaction

The overall order is the sum \( n + m = 2 + 3 = 5 \).

Key concepts

Reaction Rate

Reaction rate refers to the speed at which a chemical reaction proceeds. It tells us how quickly reactants are being converted into products. In the reaction to form ammonia (\(\mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}\),), the rate indicates how fast nitrogen and hydrogen gases come together to form ammonia. Reaction rates can be affected by several factors, including the concentrations of reactants, temperature, and the presence of catalysts.
Understanding the reaction rate is crucial as it helps chemists control the conditions to optimize the yield of a reaction. Rates are usually measured as a change in concentration per unit time, such as moles per liter per minute (\(\text{mol/L} \cdot \text{min}\)). For this specific query, by examining given experimental data, one can determine how variations in reactant concentrations affect the overall reaction rate.

Rate Equation

The rate equation expresses the relationship between the reaction rate and the concentrations of the reactants. It is written as:Rate \(= k[\text{Reactant}_1]^{n}[\text{Reactant}_2]^{m}\),where the rate is dependent on a rate constant (\(k\)) and the concentrations of the reactants raised to their respective powers (\(n\) and \(m\)). For the synthesis of ammonia, the derived rate equation is:

• Rate \(= k[\mathrm{N}_2]^2[\mathrm{H}_2]^3\)

This indicates that the rate is proportional to the concentration of nitrogen raised to the power of 2 and hydrogen raised to the power of 3. The exponents in the equation are determined experimentally and signify the reaction order with respect to each reactant. The rate equation provides insights into how the concentrations of reactants impact the speed of a reaction.

Reaction Order

Reaction order is a key concept that refers to the power to which the concentration of a reactant is raised in the rate equation. In other words, it shows the dependence of the rate on the concentration of each reactant. For the ammonia formation reaction:

• The order with respect to \(\mathrm{N}_2\) is 2.
• The order with respect to \(\mathrm{H}_2\) is 3.

This means that if the concentration of nitrogen is doubled, the rate would quadruple; if the concentration of hydrogen is doubled, the rate would increase by a factor of eight.
The overall reaction order is the sum of the orders with respect to each reactant, which in this case is:2 + 3 = 5.This helps us understand how sensitive the reaction rate is to changes in all of the reactants' concentrations.

Rate Constant

The rate constant (\(k\)) is a proportionality constant in the rate equation. It is an intrinsic part of the reaction kinetics and provides valuable insight into the reaction speed, irrespective of the concentrations of reactants. In the reaction to form ammonia, the rate constant was calculated using experimental data on concentration and rate.To calculate \(k\), we rearrange the given rate equation and substitute the known values:

• From experiment 1: the rate is given as \(4.21 \times 10^{-5}\) \(\text{mol/L} \cdot \text{min}\)
• Substituting the concentrations for \(\mathrm{N}_2\) and \(\mathrm{H}_2\), and isolating \(k\), we have\( k = \frac{4.21 \times 10^{-5}}{(0.030)^2(0.010)^3} \)

This results in \(k = 4.68 \times 10^3 \, \mathrm{L^4/mol^4/min}\).The units of \(k\) can change depending on the reaction order, and they tell us about the reaction's dependence on the concentration of the reactants. The rate constant is critical for predicting how fast a reaction will occur under given conditions.