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Ammonia is produced by the reaction between nitrogen and hydrogen gases. (a) Write a balanced equation using smallest whole-number coefficients for the reaction. (b) Write an expression for the rate of reaction in terms of \(\Delta\left[\mathrm{NH}_{3}\right]\) (c) The concentration of ammonia increases from \(0.257 \mathrm{M}\) to \(0.815 \mathrm{M}\) in \(15.0 \mathrm{~min}\). Calculate the average rate of reaction over this time interval. (d) Based on your result in part (c), calculate the rate of change of concentration of \(\mathrm{H}_{2}\) during the same time interval.

Short Answer

Expert verified
(a) \[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]; (b) \[ \frac{1}{2}\frac{\Delta[\text{NH}_3]}{\Delta t} \]; (c) 0.0372 M/min; (d) -0.0558 M/min.

Step by step solution

01

Write the Balanced Equation

To write the balanced chemical equation for the production of ammonia, we start with the reactants and products. The reaction involves nitrogen gas (\( N_2 \)) and hydrogen gas (\( H_2 \)) forming ammonia (\( NH_3 \)). The unbalanced chemical equation is:\[ N_2(g) + H_2(g) \rightarrow NH_3(g) \]Balance the equation by matching the number of each type of atom on both sides. The balanced equation is:\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
02

Expression for the Rate of Reaction

The rate of reaction can be expressed in terms of the change in concentration of ammonia (\( \Delta[\text{NH}_3] \)) over time (\( \Delta t \)). The stoichiometry of the reaction based on the balanced equation shows that for every 2 moles of \( NH_3 \) produced, 3 moles of \( H_2 \) are consumed. The rate can be written as:\[ \text{Rate} = \frac{1}{2} \frac{\Delta [\text{NH}_3]}{\Delta t} \]
03

Calculate the Average Rate of Reaction

Given that the concentration of \( NH_3 \) increases from 0.257 M to 0.815 M in 15.0 minutes, we can calculate the average rate of reaction using:\[ \text{Average rate} = \frac{\Delta [\text{NH}_3]}{\Delta t} = \frac{0.815 - 0.257}{15.0} \]Calculate:\[ \text{Average rate} = \frac{0.558}{15.0} = 0.0372 \text{ M/min} \]
04

Calculate the Rate of Change of \( H_2 \) Concentration

The rate of change of \( H_2 \) concentration can be determined from the stoichiometry of the balanced equation. From the balanced equation, 3 moles of \( H_2 \) are consumed for every 2 moles of \( NH_3 \) produced. Therefore, the rate of consumption of \( H_2 \) is:\[ \text{Rate of } \Delta [\text{H}_2] = -\frac{3}{2} \times 0.0372 \text{ M/min} \]Calculate:\[ \text{Rate of } \Delta [\text{H}_2] = -0.0558 \text{ M/min} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equation
A balanced chemical equation ensures that the number of atoms for each element is equal on both sides of the chemical reaction. This is crucial because it reflects the conservation of mass, meaning matter cannot be created or destroyed. In the reaction to form ammonia, we start with nitrogen gas \( (N_2) \) and hydrogen gas \( (H_2) \), which react to form ammonia \( (NH_3) \). The unbalanced equation initially appears as: \[ N_2(g) + H_2(g) \rightarrow NH_3(g) \] To balance the reaction, adjust the coefficients, which are the numbers before the molecules. The balanced equation becomes: \[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \] This indicates that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia.
Rate of Reaction
The rate of reaction gives us an idea of how fast or slow a reaction occurs. It is a measure of the change in concentration of reactants or products over time. In chemical kinetics, the rate can often be represented in terms of concentration changes for specific reactants or products in a given period. For ammonia production, the rate of reaction concerning ammonia can be expressed by considering how its concentration changes over time: \[ \text{Rate} = \frac{1}{2} \frac{\Delta [NH_3]}{\Delta t} \] Here, \( \Delta [NH_3] \) represents the change in ammonia concentration, and \( \Delta t \) is the time interval. The factor \( \frac{1}{2} \) arises from the stoichiometry in the balanced equation, indicating that the reaction produces 2 moles of \( NH_3 \) per reaction cycle.
Stoichiometry
Stoichiometry involves calculating the quantitative relationships between reactants and products in a chemical reaction. It derives directly from the balanced chemical equation. In the context of producing ammonia, the stoichiometry gives us the ratio of hydrogen to ammonia, which is 3:2. This means that for every three moles of hydrogen consumed, two moles of ammonia are produced. Using stoichiometry allows us to relate the rates at which different substances in a reaction are consumed or produced. Knowing the stoichiometric ratios helps calculate other quantities related to the reaction, such as the rates of consumption of other reactants like hydrogen.
Reaction Rate Calculation
Calculating the reaction rate involves determining how fast reactants are converted into products. To find the average rate of reaction for producing ammonia, use the change in concentration divided by the time interval. With the concentration of \( NH_3 \) increasing from 0.257 M to 0.815 M over 15 minutes, the calculation would be: \[ \text{Average rate} = \frac{0.815 - 0.257}{15.0} \] Simplified, this becomes: \[ \text{Average rate} = 0.0372 \text{ M/min} \] Additionally, if we want to determine the rate of hydrogen consumption, use the stoichiometric ratio from the balanced chemical equation. The rate of \( H_2 \) change is: \[ \text{Rate of } \Delta [H_2] = -\frac{3}{2} \times 0.0372 \text{ M/min} \] The negative sign indicates that \( H_2 \) is being consumed: \[ \text{Rate of } \Delta [H_2] = -0.0558 \text{ M/min} \]

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Most popular questions from this chapter

Experiments show that the reaction of nitrogen dioxide with fluorine $$ 2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{FNO}_{2}(\mathrm{~g}) $$ has the rate law $$ \text { Rate }=k\left[\mathrm{NO}_{2}\right]\left[\mathrm{F}_{2}\right] $$ and the reaction is thought to occur in two steps: $$ \begin{array}{l} \text { Step } 1: \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \\ \text { Step } 2: \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g}) \end{array} $$ (a) Show that the sum of this sequence of reactions gives the balanced equation for the overall reaction. (b) Which step is rate-determining?

The deep blue compound \(\mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}\) can be made from the chromate ion by using hydrogen peroxide in an acidic solution. \(\mathrm{HCrO}_{4}^{-}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The kinetics of this reaction have been studied, and the rate equation is Rate of disappearance of \(\mathrm{HCrO}_{4}^{-}=k\left[\mathrm{HCrO}_{4}^{-}\right]\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]\) One of the mechanisms suggested for the reaction is $$ \begin{array}{c} \mathrm{HCrO}_{4}^{-}+\mathrm{H}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{CrO}_{4} \\ \mathrm{H}_{2} \mathrm{CrO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}+\mathrm{H}_{2} \mathrm{O} \end{array} $$ \(\mathrm{H}_{2} \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (a) Give the order of the reaction with respect to each reactant. (b) Show that the steps of the mechanism agree with the overall equation for the reaction. (c) Which step in the mechanism is rate-limiting? Explain your answer.

In a time-resolved picosecond spectroscopy experiment, Sheps, Crowther, Carrier, and Crim (Journal of Physical Chemistry \(A,\) Vol. 110,\(2006 ;\) pp. \(3087-3092\) ) generated chlorine atoms in the presence of pentane. The pentane was dissolved in dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\). The chlorine atoms are free radicals and are very reactive. After a nanosecond the chlorine atoms have reacted with pentane molecules, removing a hydrogen atom to form \(\mathrm{HCl}\) and leaving behind a pentane radical with a single unpaired electron. The equation is \(\mathrm{Cl} \cdot(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{12}(\mathrm{dcm}) \longrightarrow \mathrm{HCl}(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{11} \cdot(\mathrm{dcm})\) where \((\mathrm{dcm})\) indicates that a substance is dissolved in dichloromethane. Measurements of the concentration of chlorine atoms were made as a function of time at three different concentrations of pentane in the dichloromethane. These results are shown in the table. of Chlorine Atoms (M) \begin{tabular}{clcc} & \multicolumn{3}{c} { Concentration of Chlorine Atoms (m) } \\ Time (ps) & \multicolumn{4}{c} { for Different \(\mathrm{C}_{5} \mathrm{H}_{12}\) Concentrations } \\ \cline { 3 - 5 } & \(0.15-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.30-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.60-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) \\ \hline 100.0 & \(4.42 \times 10^{-5}\) & \(3.11 \times 10^{-5}\) & \(1.77 \times 10^{-5}\) \\ 140.0 & \(3.823 \times 10^{-5}\) & \(2.39 \times 10^{-5}\) & \(1.15 \times 10^{-5}\) \\\ 180.0 & \(3.38 \times 10^{-5}\) & \(1.94 \times 10^{-5}\) & \(8.48 \times 10^{-6}\) \\\ 220.0 & \(3.03 \times 10^{-5}\) & \(1.49 \times 10^{-5}\) & \(6.04 \times 10^{-6}\) \\\ 260.0 & \(2.68 \times 10^{-5}\) & \(1.19 \times 10^{-5}\) & \(4.12 \times 10^{-6}\) \\\ 300.0 & \(2.42 \times 10^{-5}\) & \(9.45 \times 10^{-6}\) & \(3.14 \times 10^{-6}\) \\\ 340.0 & \(2.08 \times 10^{-5}\) & \(7.75 \times 10^{-6}\) & \(2.38 \times 10^{-6}\) \\\ 380.0 & \(1.91 \times 10^{-5}\) & \(6.35 \times 10^{-6}\) & \(1.75 \times 10^{-6}\) \\\ 420.0 & \(1.71 \times 10^{-5}\) & \(4.58 \times 10^{-6}\) & \(1.61 \times 10^{-6}\) \\\ 460.0 & \(1.53 \times 10^{-5}\) & \(3.77 \times 10^{-6}\) & \(9.98 \times 10^{-7}\) \\\ \hline \end{tabular} (a) Determine the order of the reaction with respect to chlorine. (b) Determine whether the reaction rate depends on the concentration of pentane in dichloromethane. If so, determine the order of the reaction with respect to pentane. (c) Explain why the concentration of pentane in dichloromethane does not affect the data analysis that you performed in part (a). (d) Write the rate law for the reaction and calculate the rate of reaction for a concentration of chlorine atoms equal to \(1.0 \mu \mathrm{M}\) and a pentane concentration of \(0.23 \mathrm{M}\). (e) Sheps, Crowther, Carrier, and Crim found that the rate of formation of \(\mathrm{HCl}\) matched the rate of disappearance of Cl. From this they concluded that there were no intermediates and side reactions were not important. Explain the ic

A cube of aluminum \(1.0 \mathrm{~cm}\) on each edge is placed into \(9-\mathrm{M} \mathrm{NaOH}(\mathrm{aq})\), and the rate at which \(\mathrm{H}_{2}\) gas is given off is measured. (a) Calculate by what factor this reaction rate will change if the aluminum cube is cut exactly in half and the two halves are placed in the solution. Assume that the reaction rate is proportional to the surface area, and that all of the surface of the aluminum is in contact with the \(\mathrm{NaOH}(\mathrm{aq})\) (b) If you had to speed up this reaction as much as you could without raising the temperature, what would you do to the aluminum?

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