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Chapter 10: Problem 108

Consider the following reaction $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ The rate of this reaction in terms of \(\mathrm{N}_{2}\) at \(\mathrm{T}\) is \(-\mathrm{d}\left[\mathrm{N}_{2}\right] /\) \(\mathrm{dt}=0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) What is the value of \(-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}\) (in units of \(\left.\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) at the same temperature? (a) \(0.02\) (b) 50 (c) \(0.06\) (d) \(0.04\)

Short Answer

Expert verified
The value is 0.06 mol L鈦宦� s鈦宦� (option c).

Step by step solution

01

Understand the Rate Expression

The reaction is balanced as follows: \[ \text{N}_{2}(\mathrm{g}) + 3\text{H}_{2}(\mathrm{g}) \longrightarrow 2\text{NH}_{3}(\mathrm{g}) \] The rate expression involving reactants tells us how fast the concentration of the reactants is decreasing. For \(\text{N}_{2}\), the rate of disappearance is given as \(-\frac{d[\text{N}_{2}]}{dt} = 0.02 \text{ mol L}^{-1} \text{ s}^{-1}\). We need to find \(-\frac{d[\text{H}_{2}]}{dt}\).
02

Use the Stoichiometry of the Reaction

According to the balanced chemical equation, 1 mol of \(\text{N}_{2}\) reacts with 3 mol of \(\text{H}_{2}\). This means that \(\text{H}_{2}\) is used up at a rate that is three times faster than \(\text{N}_{2}\). Therefore, the rate of disappearance of \(\text{H}_{2}\) can be expressed as \(-\frac{d[\text{H}_{2}]}{dt} = 3 \times (-\frac{d[\text{N}_{2}]}{dt})\).
03

Calculate \(-\frac{d[H_2]}{dt}\)

Substituting the given rate of disappearance for \(\text{N}_{2}\): \[ -\frac{d[\text{H}_{2}]}{dt} = 3 \times 0.02 \text{ mol L}^{-1} \text{ s}^{-1}\] Calculate the value: \[ -\frac{d[\text{H}_{2}]}{dt} = 0.06 \text{ mol L}^{-1} \text{ s}^{-1}\]
04

Select the Correct Answer

The calculated rate of disappearance of \(\text{H}_{2}\) is \(0.06 \text{ mol L}^{-1} \text{ s}^{-1}\). Looking at the provided options, the correct choice is \(c)\) \(0.06\).

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Key Concepts

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

Reaction Rate
The reaction rate is a measure of how quickly a chemical reaction occurs. It is typically expressed in terms of how fast the concentration of a reactant decreases or how fast the concentration of a product increases, over time. For example, in our given reaction involving nitrogen \( \text{N}_2 \\) and hydrogen \( \text{H}_2 \\), the rate is shown as a change in concentration over a change in time. \Understanding the reaction rate helps chemists to control and optimize reactions for industrial processes. It can vary based on different factors such as temperature, concentration of reactants, and presence of catalysts. \
    \
  • High reaction rates mean a reaction proceeds quickly to completion, which is beneficial in manufacturing.
    • \
  • Low reaction rates can lead to inefficiencies or incomplete reactions, undesirable in many settings.
    • \
\Overall, by understanding the rate of a reaction, we gain insights into the dynamics of the reaction and can make informed adjustments to conditions as needed.
Chemical Stoichiometry
Chemical stoichiometry gives us the quantitative relationship between reactants and products in a balanced chemical equation. It provides the ratio in which chemicals combine to form a product. In our reaction: \[ \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}) \] stoichiometry tells us that: \
    \
  • 1 mole of nitrogen gas \(\text{N}_2\\) reacts with 3 moles of hydrogen gas \(\text{H}_2\\).
    • \
  • These reactants produce 2 moles of ammonia \(\text{NH}_3\\).
    • \
\Understanding this stoichiometric ratio allows us to relate the rate of disappearance of \(\text{N}_2\) with \(\text{H}_2\\). Since for every 1 mole of \(\text{N}_2\\) consumed, 3 moles of \(\text{H}_2\\) are consumed, this relationship is crucial for calculating reaction rates based on one component when another is known. \Stoichiometry is a fundamental concept that helps chemists predict the quantities of substances consumed and produced in a reaction, ensuring precise and efficient use of materials in both laboratory and industrial processes.
Rate of Reaction
The rate of reaction specifies how fast reactants are converted into products. In our example, the rate has been given for nitrogen \(\text{N}_2\\) as \(-\frac{d[\text{N}_2]}{dt} = 0.02\ \text{mol L}^{-1} \text{s}^{-1}\). This tells us how quickly \(\text{N}_2\\) is being consumed. \To find the rate of reaction for \(\text{H}_2\\), which is consumed at a different rate due to stoichiometry, we apply the stoichiometric relationship discussed earlier:
  • For every mole of \(\text{N}_2\\) consumed, 3 moles of \(\text{H}_2\\) are consumed.
    • \
\Therefore, the rate \(-\frac{d[\text{H}_2]}{dt}\)\ is 3 times that of \(-\frac{d[\text{N}_2]}{dt}\), which calculates to \(0.06 \ \text{mol L}^{-1} \text{s}^{-1}\). \By understanding the rate of reaction, scientists can adjust conditions to speed up slow reactions or control fast reactions, which is essential for efficiency in chemical production and safety.

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Most popular questions from this chapter

\(3 \mathrm{~A} \longrightarrow 2 \mathrm{~B}\), rate of reaction \(+\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\) is equal to (a) \(-\frac{3}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) (b) \(-\frac{2}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) (c) \(-\frac{1}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) (d) \(+2 \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)

The following data pertains to the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) $$ \begin{array}{llll} \hline \text { S. No. } & {[\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1}} & {[\mathrm{~B}] \mathrm{mol} \mathrm{L}^{-1}} & \begin{array}{l} \text { Rate mol } \\ \mathrm{L}-\mathbf{S}^{-1} \end{array} \\ \hline \text { 1. } & 1 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-4} \\\ 2 . & 2 \times 10^{-2} & 2 \times 10^{-2} & 4 \times 10^{-4} \\ 3 . & 2 \times 10^{-2} & 4 \times 10^{-2} & 8 \times 10^{-8} \\ \hline \end{array} $$ Which of the following inferences are drawn from the above data? (1) rate constant of the reaction is \(10^{-4}\) (2) rate law of the reaction is \([\mathrm{A}][\mathrm{B}]\) (3) rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: (a) 1 and 3 (b) 2 and 3 (c) land 2 (d) 1,2 and 3

The experimental data for the reaction \(2 \mathrm{~A}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}\) is $$ \begin{array}{llll} \hline \text { Exp. } & {[\mathrm{A}]} & {\left[\mathrm{B}_{2}\right]} & \text { Rate }\left(\mathrm{Ms}^{-1}\right) \\ \hline 1 . & 0.50 \mathrm{M} & 0.50 \mathrm{M} & 1.6 \times 10^{-4} \\ 2 . & 0.50 \mathrm{M} & 1.00 \mathrm{M} & 3.2 \times 10^{-4} \\ 3 . & 1.00 \mathrm{M} & 1.00 \mathrm{M} & 3.2 \times 10^{-4} \\ \hline \end{array} $$ the rate equation for the above data is (a) rate \(=\mathrm{k}\left[\mathrm{B}_{2}\right]\) (b) rate \(=k\left[\mathrm{~B}_{2}\right]^{2}\) (c) rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{2}\) (d) rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]\)

For a zero-order reaction, the plot of concentration vs time is linear with (a) \(+\) ve slope and zero intercept (b) -ve slope and zero intercept (c) tve slope and non-zero intercept (d) -ve slope and non-zero intercept

At \(380^{\circ} \mathrm{C}\), half-life period for the first-order decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(360 \mathrm{~min}\). The energy of activation of the reaction is \(200 \mathrm{~kJ} \mathrm{~mol}^{1} .\) Calculate the time required for \(75 \%\) decomposition at \(450^{\circ} \mathrm{C}\) if half-life for decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(10.17 \mathrm{~min}\) at \(450^{\circ} \mathrm{C}\). (a) \(20.4 \mathrm{~min}\) (b) \(408 \mathrm{~min}\) (c) \(10.2 \mathrm{~min}\) (d) none of these
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