Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In a study of the formation of HI from its elements, $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ equal amounts of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) were placed in a container, which was then sealed and heated. (a) On one set of axes, sketch concentration vs. time curves for \(\mathrm{H}_{2}\) and HI, and explain how \(Q\) changes as a function of time. (b) Is the value of \(Q\) different if \(\left[\mathrm{I}_{2}\right]\) is plotted instead of \(\left[\mathrm{H}_{2}\right] ?\)

Short Answer

Expert verified
In the reaction \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g)\), the value of \(Q\) increases over time until it equals the equilibrium constant \(K\). The value of \(Q\) is the same when \([I_2]\) is plotted instead of \([H_2]\).

Step by step solution

01

Understanding the Reaction

The given reaction is \ \[ \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \]. \ It indicates that hydrogen gas and iodine gas react to form hydrogen iodide gas.
02

Sketching the Concentration vs. Time Curves

On a set of axes, plot concentration (vertical axis) against time (horizontal axis). \[ \text{For } \mathrm{H}_{2}: \text{Start at } [H_2]_0 \text{ and decrease over time until equilibrium is reached, forming a curve.} \] \[ \text{For } \mathrm{HI}: \text{Start at 0 and increase as } HI \text{ is formed, then level off at equilibrium.} \]
03

Explaining the Changes in Q

The reaction quotient (\(Q\)) is defined as \(Q = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\). Initially, \(Q\) is zero because \( \left[ \mathrm{HI} \right] = 0 \). \ As the reaction proceeds, \(Q\) increases as the concentration of \( \mathrm{HI} \) increases and the concentrations of \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \) decrease until equilibrium is reached. At equilibrium, \(Q = K \), where \(K\) is the equilibrium constant.
04

Analyzing the Plot of [I2]

If the concentration of \( \mathrm{I}_2 \) is plotted instead of \( \mathrm{H}_2 \), the value of \(Q\) remains the same because \( \left[ \mathrm{I}_2 \right] \) is directly proportional to \( \left[ \mathrm{H}_2 \right] \) due to the stoichiometry of the reaction.\

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

reaction quotient
The reaction quotient, or \(Q\), is a measure used to determine the direction in which a reaction will proceed to reach equilibrium. For the reaction involving \(\text{H}_2\) and \(\text{I}_2\) to form \(2 \text{HI}\), \(Q\) is calculated using the formula: \( Q = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \). The values of concentrations are taken from the instant you are interested in.
For example, at the start of the reaction, when no product has formed, \([ \text{HI} ] = 0\), making \(Q = 0\). As the reaction progresses, the concentration of \(\text{HI}\) increases while that of \(\text{H}_2\) and \(\text{I}_2\) decreases. This makes \(Q\) increase over time until it matches the equilibrium constant \(K\), indicating that the equilibrium has been reached.
equilibrium constant
The equilibrium constant, represented as \(K\), is a fixed value that describes the ratio of product concentrations to reactant concentrations at equilibrium for a reversible reaction. For our reaction \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g)\), \(K\) is given by \(K = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\).
It's crucial to note that \(K\) remains constant for a given reaction at a specific temperature. It does not change even if the initial concentrations of the reactants or products are altered. At equilibrium, the reaction quotient \(Q\) reaches this constant value \(K\), indicating that the reaction mixtures are in a state of balance where the rate of the forward reaction equals the rate of the reverse reaction.
concentration vs. time curves
Concentration vs. time curves offer a visual representation of how the concentrations of reactants and products change over time. For the reaction involving \(\text{H}_2\) and \(\text{I}_2\) to form \(2 \text{HI}\), these curves can be plotted on a graph with concentration on the vertical axis and time on the horizontal axis.
Initially, the concentration of \(\text{H}_2\) and \(\text{I}_2\) starts at their initial values, while the concentration of \(\text{HI}\) starts at zero. As the reaction proceeds:
  • The concentrations of \(\text{H}_2\) and \(\text{I}_2\) decrease as they are consumed.
  • The concentration of \(\text{HI}\) increases as it forms, eventually flattening out as equilibrium is approached.
These curves provide insight into how quickly a reaction progresses and when equilibrium is reached.
stoichiometry
Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction, based on their balanced chemical equation. For the reaction \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g)\), the stoichiometric coefficients are 1:1:2, meaning one molecule of \(\text{H}_2\) reacts with one molecule of \(\text{I}_2\) to produce two molecules of \(\text{HI}\).
This stoichiometric relationship helps determine how the concentrations of the reactants and product will change over time. For example:
  • If 1 mole of \(\text{H}_2\) reacts fully, it would need 1 mole of \(\text{I}_2\), and together they would produce 2 moles of \(\text{HI}\).
  • The stoichiometry also ensures that the decrease in the concentrations of \(\text{H}_2\) and \(\text{I}_2\) is proportional to their coefficients, maintaining balance in the reaction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For a problem involving the catalyzed reaction of methane and steam, the following reaction table was prepared: $$ \begin{array}{lcccc} \begin{array}{l} \text { Pressure } \\ \text { (atm) } \end{array} & \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(g) & \Longrightarrow & \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g) \\ \hline \text { Initial } & 0.30 & 0.40 & 0 & 0 \\ \text { Change } & -x & -2 x & +x & +4 x \\ \hline \text { Equilibrium } & 0.30-x & 0.40-2 x & x & 4 x \end{array} $$ Explain the entries in the "Change" and "Equilibrium" rows.

Balance each of the following examples of heterogeneous equilibria and write each reaction quotient, \(Q_{\mathrm{c}}\) : (a) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{O}_{2}(g)\) (b) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\)

At \(100^{\circ} \mathrm{C}, K_{\mathrm{p}}=60.6\) for the reaction $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ In a given experiment, 0.10 atm of each component is placed in a container. Is the system at equilibrium? If not, in which direction will the reaction proceed?

As an EPA scientist studying catalytic converters and urban smog, you want to find \(K_{c}\) for the following reaction: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \quad K_{\mathrm{c}}=? $$ Use the following data to find the unknown \(K_{\mathrm{c}}\) : $$ \begin{aligned} \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \rightleftharpoons \mathrm{NO}(g) & & K_{\mathrm{c}}=4.8 \times 10^{-10} \\ 2 \mathrm{NO}_{2}(g) & \Longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & & K_{\mathrm{c}}=1.1 \times 10^{-5} \end{aligned} $$

Predict the effect of increasing the container volume on the amounts of each reactant and product in the following reactions: (a) \(\mathrm{CH}_{3} \mathrm{OH}(l) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) (b) \(\mathrm{CH}_{4}(g)+\mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{HCN}(g)+3 \mathrm{H}_{2}(g)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free