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Chapter 13: Problem 40

For the reaction of hydrogen with iodine $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g) $$ relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide.

Short Answer

Expert verified
-d[H鈧俔/dt = (1/2)d[HI]/dt.

Step by step solution

01

Understand the Reaction

The given reaction is a simple combination reaction where one mole of hydrogen gas reacts with one mole of iodine gas to produce two moles of hydrogen iodide gas. It is important to note that the stoichiometry of the reaction is 1:1:2.
02

Write the Expression for Rate of Disappearance

The rate of disappearance of a reactant refers to how quickly it is used up as the reaction occurs. For hydrogen gas, we write the rate of disappearance as \[-\frac{d[ ext{H}_2]}{dt}\].
03

Write the Expression for Rate of Formation

The rate of formation of a product refers to how quickly it is formed during the reaction. For hydrogen iodide gas, the rate of formation is written as \[\frac{d[ ext{HI}]}{dt}\].
04

Relate the Rates Using Stoichiometry

From the balanced chemical equation, we know that 1 mole of \( ext{H}_2\) produces 2 moles of \( ext{HI}\). Thus, the rate of disappearance of \( ext{H}_2\) is half of the rate of formation of \( ext{HI}\), which can be mathematically expressed as:\[-\frac{1}{1}\frac{d[ ext{H}_2]}{dt} = \frac{1}{2}\frac{d[ ext{HI}]}{dt}\].
05

Final Equation for Relation

The relation can be simplified to \[- \frac{d[ ext{H}_2]}{dt} = \frac{1}{2}\frac{d[ ext{HI}]}{dt}\]. This equation shows that for every mole of \( ext{H}_2\) that disappears, two moles of \( ext{HI}\) are formed.

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Key Concepts

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

Rate of Disappearance
In the realm of reaction kinetics, the rate of disappearance refers to the speed at which a reactant is consumed over time in a chemical reaction. For hydrogen gas (\(\text{H}_2\)) reacting with iodine (\(\text{I}_2\)), its rate of disappearance is an essential concept. It tells us how fast hydrogen is being used up in the reaction.
  • The rate is mathematically expressed as a negative derivative since the concentration of hydrogen decreases: \[-\frac{d[\text{H}_2]}{dt}\].
  • The negative sign indicates a reduction in concentration of \(\text{H}_2\) over time.
Understanding this rate helps chemists predict how quickly the reactants are being utilized. This is crucial for controlling reaction conditions in industrial applications or even in laboratory experiments. By knowing how fast a reactant disappears, adjustments can be made to optimize the efficiency and yield of the desired products.
Rate of Formation
The rate of formation is another key component in understanding how reactions proceed. It describes how quickly a product forms from the reactants as the reaction progresses. In our equation, the compound of interest is hydrogen iodide (\(\text{HI}\)).
  • The expression for the rate of formation is given as: \[\frac{d[\text{HI}]}{dt}\].
  • This positive rate indicates that the concentration of hydrogen iodide is increasing over time.
By examining the rate of formation, chemists can gain insights into how efficiently a reaction is producing its intended products. This aspect is especially significant when scaling reactions for manufacturing processes where the timely production of a compound can affect the profitability and feasibility of the process.
Chemical Stoichiometry
Chemical stoichiometry is a fundamental concept that relates the quantities of reactants and products in a balanced chemical reaction. Through stoichiometry, we understand the proportional relationships between reactants and products. In our given reaction \(\text{H}_2\) + \(\text{I}_2\) \(\rightarrow\) 2 \(\text{HI}\), stoichiometry plays a vital role.
  • The balanced equation tells us that 1 mole of hydrogen gas reacts with 1 mole of iodine gas to produce 2 moles of hydrogen iodide gas.
  • This relationship allows us to express rates: for every 1 mole of hydrogen that disappears, 2 moles of hydrogen iodide are formed. Therefore, the stoichiometric relationship is used to relate the rates, as seen in the equation:\[- \frac{d[\text{H}_2]}{dt} = \frac{1}{2}\frac{d[\text{HI}]}{dt}\].
Understanding stoichiometry is crucial as it helps in the calculation of how much reactant is needed or how much product will be produced in a reaction. This knowledge is essential for both theoretical calculations and practical applications in chemical reactions.

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Most popular questions from this chapter

A study of the decomposition of azomethane, \(\mathrm{CH}_{3} \mathrm{NNCH}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{N}_{2}(g)\) gave the following concentrations of azomethane at various times: \(\begin{array}{rl}\text { Time } & {\left[\mathrm{CH}_{3} \mathrm{NNCH}_{3}\right]} \\ 0 \mathrm{~min} & 1.50 \times 10^{-2} \mathrm{M} \\\ 10 \mathrm{~min} & 1.29 \times 10^{-2} \mathrm{M} \\ & 1.10 \times 10^{-2} \mathrm{M}\end{array}\) \(20 \mathrm{~min} \quad 1.10 \times 10^{-2} M\) $$ 30 \mathrm{~min} \quad 0.95 \times 10^{-2} \mathrm{M} $$ Obtain the average rate of decomposition in units of \(M / \mathrm{s}\) for each time interval.

In the presence of excess thiocyanate ion, \(\mathrm{SCN}^{-}\), the following reaction is first order in iron(III) ion, \(\mathrm{Fe}^{3+} ;\) the rate constant is \(1.27 / \mathrm{s}\). $$ \mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \longrightarrow \mathrm{Fe}(\mathrm{SCN})^{2+}(a q) $$ What is the half-life in seconds? How many seconds would be required for the initial concentration of \(\mathrm{Fe}^{3+}\) to decrease to each of the following values: \(25.0 \%\) left, \(12.5 \%\) left, \(6.25 \%\) left, \(3.125 \%\) left? What is the relationship between these times and the half-life?

Draw a potential-energy diagram for an uncatalyzed exothermic reaction. On the same diagram, indicate the change that results on the addition of a catalyst. Discuss the role of a catalyst in changing the rate of reaction.

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), undergoes first-order decomposition in chloroform solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The rate constant at \(45^{\circ} \mathrm{C}\) is \(6.2 \times 10^{-4} / \mathrm{min} .\) Calculate the volume of \(\mathrm{O}_{2}\) obtained from the reaction of \(1.00 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\) at \(45^{\circ} \mathrm{C}\) and \(770 \mathrm{mmHg}\) after 20.0 hr.

The reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) is found to be zero order. If it takes \(5.2 \times 10^{2}\) seconds for an initial concentration of A to go from \(0.50 M\) to \(0.25 M\), what is the rate constant for the reaction?
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