/*! 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 34 Equal numbers of moles of \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Equal numbers of moles of \(\mathrm{H}_{2}\) gas and \(\mathrm{I}_{2}\) vapor are mixed in a flask and heated to \(700^{\circ}\) C. The initial concentration of each gas is \(0.0088 \mathrm{mol} / \mathrm{L},\) and \(78.6 \%\) of the \(\mathrm{I}_{2}\) is consumed when equilibrium is achieved according to the equation $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{HI}(\mathrm{g}) $$ Calculate \(K\) for this reaction.

Short Answer

Expert verified
The equilibrium constant \( K \) is approximately 55.00.

Step by step solution

01

Determine Change in Concentration

When the reaction reaches equilibrium, we know that 78.6% of \( \mathrm{I_2} \) is consumed. The initial concentration of \( \mathrm{I_2} \) is \( 0.0088 \, \text{mol/L} \). Thus, the change in concentration \( \Delta c \) is given by \( 0.0088 \, \text{mol/L} \times 0.786 = 0.00693 \, \text{mol/L} \).
02

Calculate Equilibrium Concentrations

Because \( \mathrm{H_2} \) and \( \mathrm{I_2} \) react in a 1:1 ratio to form \( \mathrm{HI} \), the change in concentration of \( \mathrm{H_2} \) is also \( 0.00693 \, \text{mol/L} \). At equilibrium, the concentration of \( \mathrm{H_2} \) and \( \mathrm{I_2} \) is \( 0.0088 - 0.00693 = 0.00187 \, \text{mol/L} \). The change leads to an increase of \( 2 \times 0.00693 = 0.01386 \, \text{mol/L} \) for \( \mathrm{HI} \), giving an equilibrium concentration of \( 0.01386 \, \text{mol/L} \).
03

Write and Substitute into Expression for Equilibrium Constant

The equilibrium constant expression for the reaction is: \[K = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}\]Substituting the equilibrium concentrations into this equation, we have: \[K = \frac{(0.01386)^2}{(0.00187)(0.00187)}\]
04

Calculate the Equilibrium Constant \( K \)

Perform the calculation: \[K = \frac{(0.01386)^2}{(0.00187)(0.00187)} = \frac{0.0001922996}{0.0000034969} \approx 55.00\]Therefore, the equilibrium constant \( K \) is approximately \( 55.00 \).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a dynamic state in which the rate of the forward reaction equals the rate of the reverse reaction. At this point, there is no net change in the concentrations of reactants and products. When we deal with a chemical equation like \[ \mathrm{H}_{2}(\mathrm{g}) + \mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{HI}(\mathrm{g}) \]it reaches equilibrium when the concentrations of \( \mathrm{H_2} \), \( \mathrm{I_2} \), and \( \mathrm{HI} \) remain constant over time. This does not mean the reactions have stopped but that the forward and backward reactions occur at the same rate.
An essential feature of equilibrium is that it can be disturbed by changing the conditions in a process known as Le Chatelier's Principle. If you change the concentration of reactants or products, or alter the temperature or pressure, the system will adjust to counteract that change and reestablish equilibrium. Understanding these principles helps in predicting the direction in which a reaction will proceed when conditions are changed.
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. In the given reaction equation, \( \mathrm{H}_{2}(\mathrm{g}) + \mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{HI}(\mathrm{g}) \), the stoichiometry shows that one mole of \( \mathrm{H}_2 \) reacts with one mole of \( \mathrm{I}_2 \) to produce two moles of \( \mathrm{HI} \). This 1:1:2 stoichiometric ratio is crucial for calculating changes in concentrations as the reaction proceeds.
In our specific case, because 78.6% of \( \mathrm{I}_2 \) is consumed, an equal amount of \( \mathrm{H}_2 \) must also be consumed. Since two moles of \( \mathrm{HI} \) are produced for every mole of \( \mathrm{I}_2 \) consumed, the amount of \( \mathrm{HI} \) formed can be calculated by doubling the change in \( \mathrm{I}_2 \). This exact ratio is vital for balancing chemical equations and performing calculations related to the yield of products.
Concentration Changes
Concentration changes play a central role in understanding chemical reactions at equilibrium. The initial concentrations of \( \mathrm{H}_{2} \) and \( \mathrm{I}_{2} \) were both \( 0.0088 \ \text{mol/L} \). With 78.6% of each being used, the concentration change, denoted \( \Delta c \), was \( 0.00693 \ \text{mol/L} \).
Using those changes, we determine equilibrium concentrations by subtracting \( \Delta c \) from the initial concentrations of \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \), resulting in \( 0.00187 \ \text{mol/L} \) for each. The concentration of \( \mathrm{HI} \) increased by \( 2 \times 0.00693 = 0.01386 \ \text{mol/L} \).
Understanding how to calculate these changes is important for applying the equilibrium constant expression \[K = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} \]. This equation links concentration changes to the equilibrium constant \( K \), providing insight into the extent of reaction and stability of the substances at equilibrium.

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Most popular questions from this chapter

Heating a metal carbonate leads to decomposition. $$ \mathrm{BaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{BaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ Predict the effect on the equilibrium of each change listed below. Answer by choosing (i) no change, (ii) shifts left, or (iii) shifts right. (a) add \(\mathrm{BaCO}_{3}\) (c) add BaO (b) add \(\mathrm{CO}_{2}\) (d) raise the temperature (e) increase the volume of the flask containing the reaction

The equilibrium constant, \(K_{\mathrm{p}}\), is 0.15 at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g}) $$ If the total pressure of the gas mixture is 2.5 atm at equilibrium, what is the partial pressure of each gas?

Summary and Conceptual Questions The following questions may use concepts from preceding chapters. Decide whether each of the following statements is true or false. If false, change the wording to make it true. (a) The magnitude of the equilibrium constant is always independent of temperature. (b) When two chemical equations are added to give a net equation, the equilibrium constant for the net equation is the product of the equilibrium constants of the summed equations. (c) The equilibrium constant for a reaction has the same value as \(K\) for the reverse reaction. (d) Only the concentration of \(\mathrm{CO}_{2}\) appears in the equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\) (e) For the reaction \(\operatorname{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\) the value of \(K\) is numerically the same no matter whether the amount of \(\mathrm{CO}_{2}\) is expressed as moles/liter or as gas pressure.

A sample of liquid water is sealed in a container. Over time some of the liquid evaporates, and equilibrium is reached eventually. At this point you can measure the equilibrium vapor pressure of the water. Is the process \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftarrows\) \(\mathrm{H}_{2} \mathrm{O}(\ell)\) a dynamic equilibrium? Explain the changes that take place in reaching equilibrium in terms of the rates of the competing processes of evaporation and condensation.

Disturbing a Chemical Equilibrium Dinitrogen trioxide decomposes to NO and \(\mathrm{NO}_{2}\) in an endothermic process \((\Delta H=40.5 \mathrm{kJ} / \mathrm{mol})\) $$ \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) $$ Predict the effect of the following changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each of the following changes is made. (a) adding more \(\mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g})\) (b) adding more \(\mathrm{NO}_{2}(\mathrm{g})\) (c) increasing the volume of the reaction flask (d) lowering the temperature

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