/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 The gas-phase reaction between h... [FREE SOLUTION] | 91Ó°ÊÓ

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The gas-phase reaction between hydrogen and iodine $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$$ proceeds with a rate constant for the forward reaction at \(700^{\circ} \mathrm{C}\) of \(138 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) and an activation energy of \(165 \mathrm{~kJ} / \mathrm{mol}\). (a) Calculate the activation energy of the reverse reaction given that \(\Delta H_{i}^{\circ}\) for HI is \(26.48 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta H_{i}^{\circ}\) for \(\mathrm{I}_{2}(\mathrm{~g})\) is \(62.44 \mathrm{~kJ} / \mathrm{mol}\). (b) Calculate the rate constant for the reverse reaction at \(700^{\circ} \mathrm{C}\). (Assume \(\mathrm{A}\) in the equation \(k=\mathrm{Ae}^{-\mathcal{P}_{2} / \mathrm{RT}^{\prime}}\) is the same for both forward and reverse reactions.) (c) Calculate the rate of the reverse reaction if the concentration of HI is \(0.200 M\). The reverse reaction is second-order in HI.

Short Answer

Expert verified
To summarize: 1. The activation energy of the reverse reaction is 174.48 kJ/mol. 2. The rate constant for the reverse reaction at 700°C is 772.7 L/mol·s. 3. The rate of the reverse reaction at a HI concentration of 0.200 M is 30.91 M/s.

Step by step solution

01

Analyze the given information

We are given the balance chemical equation, activation energy for the forward reaction \((E_{a}^{fwd})\), rate constant for the forward reaction \((k_{fwd})\), and enthalpies for HI and I2. We can use the following equation to calculate the activation energy for the reverse reaction \((E_{a}^{rev})\): \(E_{a}^{rev} = E_{a}^{fwd} - \Delta H^{\circ}\) Here, \(\Delta H^{\circ}\) is the enthalpy change of the reaction.
02

Calculate enthalpy change of reaction

Using the enthalpy values given, we can calculate the enthalpy change of the reaction: \(\Delta H^{\circ} = \Delta H_{i}^{\circ}(\mathrm{HI})\times2 - \Delta H_{i}^{\circ}(\mathrm{H_2}) - \Delta H_{i}^{\circ}(\mathrm{I_2}) = 2(26.48)-0-62.44 = -9.48\,\mathrm{kJ/mol}\)
03

Calculate activation energy of reverse reaction

Using the enthalpy change of the reaction calculated and the given activation energy for the forward reaction: \(E_{a}^{rev} = E_{a}^{fwd} - \Delta H^{\circ} = 165-(-9.48) = 174.48\,\mathrm{kJ/mol}\) 2. Calculate the rate constant for the reverse reaction at \(700^{\circ}\mathrm{C}\)
04

Use the Arrhenius equation

The Arrhenius equation is given as: \(k=\mathrm{Ae}^{-E_{a}/\mathrm{R} T}\), where A is the pre-exponential factor, R is the gas constant, and T is the temperature. We are given that A is the same for both forward and reverse reactions. Thus, we can use this information to find the rate constant of the reverse reaction \((k_{rev})\). First, we convert the temperature to Kelvin: \(T = 700^{\circ} \mathrm{C} + 273.15 = 973.15\,\mathrm{K}\)
05

Calculate the rate constant for the reverse reaction

Using the given rate constant for the forward reaction and the calculated activation energies of forward and reverse reactions, we can find the rate constant for the reverse reaction as follows: \({k_{rev}}/{k_{fwd}} =\mathrm{e}^{-E_{a}^{rev}/\mathrm{R} T}/\mathrm{e}^{-E_{a}^{fwd}/\mathrm{R}T}\) \(k_{rev} = k_{fwd}\cdot \frac{\mathrm{e}^{-E_{a}^{rev}/\mathrm{R} T}}{\mathrm{e}^{-E_{a}^{fwd}/\mathrm{R}T}} = 138\cdot \frac{\mathrm{e}^{-174480 /8.314\times973.15}}{\mathrm{e}^{-165000 / 8.314\times973.15}}\) \(k_{rev} = 772.7\,\mathrm{L/mol\cdot s}\) 3. Calculate the rate of the reverse reaction if the concentration of HI is \(0.200\,\mathrm{M}\)
06

Use the rate law for second-order reactions

The reverse reaction is second-order in HI. The rate law for the reverse reaction is as follows: \(rate_{rev} = k_{rev}\cdot [\mathrm{HI}]^2\)
07

Calculate the rate of the reverse reaction

Using the calculated rate constant for the reverse reaction and the given concentration of HI, we calculate the rate of the reverse reaction: \(rate_{rev} = 772.7\cdot (0.200)^2\) \(rate_{rev} = 30.91\,\mathrm{M/s}\)

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Key Concepts

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

Understanding Chemical Kinetics
Chemical kinetics, the branch of physical chemistry involved in understanding the rates of chemical reactions, is crucial for the study of reaction mechanisms and the calculation of activation energy. It involves the analysis of how different experimental conditions can influence the speed of a chemical reaction and how this can be quantified through rate constants.

For a reaction such as \(\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\), knowing the rate constant for the forward reaction allows us to predict how quickly reactants will convert to products under certain conditions. Activation energy (\(E_a\)) is a vital concept in this field, representing the minimum energy that must be overcome for reactants to transform into products.

When calculating activation energy for a reverse reaction, one must recognize the role of enthalpy change (\(\Delta H^\circ\)) in the process. This change, which can be endothermic or exothermic, will either increase or decrease the overall activation energy needed for the reverse reaction compared to the forward reaction, respectively.
Exploring the Arrhenius Equation
The Arrhenius equation is a formula that describes how the rate constant (\(k\)) of a reaction varies with temperature and activation energy. The equation is expressed as \(k = Ae^{(-E_a)/(RT)}\), where \(A\) is the pre-exponential factor, a constant that represents the frequency of collisions with the correct orientation; \(E_a\) is the activation energy; \(R\) is the universal gas constant; and \(T\) is the temperature in Kelvin.

This equation helps chemists understand the sensitivity of reaction rates to temperature changes. For example, in the exercise, we're informed the pre-exponential factor is the same for both the forward and reverse reactions, allowing us to make comparisons and calculate the rate constant for the reverse reaction when given the activation energy and temperature. After converting Celsius to Kelvin, the equation enables precise calculation. Understanding and applying the Arrhenius equation are fundamental when dealing with temperature's effect on reaction rates.
Enthalpy Change in Reactions
Enthalpy change (\(\Delta H^\circ\)) in a chemical reaction refers to the heat energy absorbed or released when a reaction occurs at constant pressure. It's a state function, meaning it depends only on the initial and final states of the system, not the path taken.

The enthalpy change for a reaction can be calculated using the standard enthalpies of formation of the reactants and products. This value demonstrates whether the reaction is endothermic (heat absorbed, positive \(\Delta H^\circ\)) or exothermic (heat released, negative \(\Delta H^\circ\)).

For instance, in the provided exercise, the enthalpy change affects the activation energy of the reverse reaction. The reaction's calculated \(\Delta H^\circ\) is subtracted from the activation energy of the forward reaction to determine the activation energy for the reverse reaction according to the equation \(E_{a}^{rev} = E_{a}^{fwd} - \Delta H^\circ\). An accurate understanding of \(\Delta H^\circ\) and its implications on activation energy are essential components of reaction chemistry.

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Most popular questions from this chapter

Dinitrogen pentoxide gas decomposes to form nitrogen dioxide and oxygen. The reaction is first-order and has a rate constant of \(0.247 \mathrm{~h}^{-1}\) at \(25^{\circ} \mathrm{C}\). If a 2.50-L flask originally contains \(\mathrm{N}_{2} \mathrm{O}_{5}\) at a pressure of \(756 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\), then how many moles of \(\mathrm{O}_{2}\) are formed after 135 minutes? (Hint: First write a balanced equation for the decomposition.)

The decomposition of dimethyl ether \(\left(\mathrm{CH}_{3} \mathrm{OCH}_{3}\right)\) to methane, carbon monoxide, and hydrogen gases is found to be first-order. At \(500^{\circ} \mathrm{C}\), a \(150.0\) -mg 35 sample of dimethyl ether is reduced to \(43.2 \mathrm{mg}\) after three quarters of an hour. Calculate (a) the rate constant. (b) the half-life at \(500^{\circ} \mathrm{C}\). (c) how long it will take to decompose \(95 \%\) of the dimethyl ether.

A reaction has two reactants \(\mathrm{A}\) and \(\mathrm{B}\). What is the order with respect to each reactant and the overall order of the reaction described by each of the following rate expressions? (a) rate \(=k_{1}[\mathrm{~A}]^{3}\) (b) rate \(=k_{2}[\mathrm{~A}] \times[\mathrm{B}]\) (c) rate \(=k_{3}[\mathrm{~A}] \times[\mathrm{B}]^{2}\) (d) rate \(=k_{4}[\mathrm{~B}]\)

For the reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}$$ the rate expression is rate \(=k[\mathrm{~A}][\mathrm{B}]\) (a) Given three test tubes, with different concentrations of \(\mathrm{A}\) and \(\mathrm{B}\), which test tube has the smallest rate? (1) \(0.10 M \mathrm{~A}_{\mathrm{i}} 0.10 \mathrm{M} \mathrm{B}\) (2) \(0.15 \mathrm{MA}_{;} 0.15 \mathrm{M} \mathrm{B}\) (3) \(0.06 M \mathrm{~A} ; 1.0 M \mathrm{~B}\) (b) If the temperature is increased, describe (using the words increases, decreases, or remains the same) what happens to the rate, the value of \(k\), and \(E_{x}\)

Two mechanisms are proposed for the reaction $$\begin{array}{cl}2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \\ \text { Mechanism 1: } \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3} & \text { (fast) } \\ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} & \text { (slow) } \\ \text { Mechanism 2: } \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} & \text { (fast) } \\ \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} & \text { (slow) } \end{array}$$ Show that each of these mechanisms is consistent with the observed rate law: rate \(=k[\mathrm{NO}]^{2} \times\left[\mathrm{O}_{2}\right]\).

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