/*! 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 79 The first step in the industrial... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The first step in the industrial synthesis of hydrogen is the reaction of steam and methane to give synthesis gas, a mixture of carbon monoxide and hydrogen: \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CH}_{4}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \quad K_{c}=4.7\) at \(1400 \mathrm{~K}\) A mixture of reactants and products at \(1400 \mathrm{~K}\) contains \(0.035\) \(\mathrm{M} \mathrm{H}_{2} \mathrm{O}, 0.050 \mathrm{M} \mathrm{CH}_{4}, 0.15 \mathrm{M} \mathrm{CO}\), and \(0.20 \mathrm{M} \mathrm{H}_{2}\). In which di- rection does the reaction proceed to reach equilibrium?

Short Answer

Expert verified
The reaction proceeds in the forward direction to reach equilibrium.

Step by step solution

01

Identify the reaction and the equilibrium constant

The reaction given is \( \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CH}_{4}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \) with an equilibrium constant \( K_c = 4.7 \) at 1400 K. This is the key parameter that will determine the position of equilibrium.
02

Write the expression for the reaction quotient \(Q_c\)

The reaction quotient, \(Q_c\), for the reaction can be expressed as \[ Q_c = \frac{[\mathrm{CO}][\mathrm{H}_2]^3}{[\mathrm{H}_2\mathrm{O}][\mathrm{CH}_4]} \] This will help us compare with \(K_c\) to determine the direction of the reaction.
03

Plug the provided concentrations into the \(Q_c\) formula

Substitute the given concentrations into the \(Q_c\) formula:\[ Q_c = \frac{(0.15)(0.20)^3}{(0.035)(0.050)} \] Calculate \(Q_c\) using these values.
04

Calculate \(Q_c\) and compare it with \(K_c\)

First, calculate \( Q_c \): - Numerator: \((0.15)(0.20)^3 = 0.15 \times 0.008 = 0.0012\)- Denominator: \((0.035)(0.050) = 0.00175\)So, \( Q_c = \frac{0.0012}{0.00175} \approx 0.686\).Compare \( Q_c \) with \( K_c = 4.7 \).
05

Determine the direction of the reaction

Given \( Q_c = 0.686 \) and \( K_c = 4.7 \), since \( Q_c < K_c \), the reaction will proceed in the forward direction to reach equilibrium. This means more products will form until equilibrium is reached.

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 91Ó°ÊÓ!

Key Concepts

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

Reaction Quotient \(Q_c\)
The reaction quotient, denoted as \( Q_c \), is a valuable tool in chemical equilibrium calculations. It is similar to the equilibrium constant \( K_c \) but can be calculated at any point during the reaction. By using \( Q_c \), we can assess the progress of the reaction and predict its direction relative to equilibrium.
In our specific reaction between steam and methane to form carbon monoxide and hydrogen, the formula for \( Q_c \) is:
  • \( Q_c = \frac{[\mathrm{CO}][\mathrm{H}_2]^3}{[\mathrm{H}_2\mathrm{O}][\mathrm{CH}_4]} \)
Plugging in the current concentrations of the reaction components tells us how the reaction is behaving at a specific moment. This helps determine whether the reaction needs to proceed forward or backward to reach equilibrium.
Equilibrium Constant \(K_c\)
The equilibrium constant \( K_c \) is a specific value that indicates the ratio of the concentrations of products to reactants at equilibrium for a chemical reaction at a given temperature. It is a fundamental concept in understanding chemical reactions.
For our industrial synthesis of hydrogen, \( K_c \) is given as 4.7 at 1400 K. This value is fixed for that specific reaction and temperature. It provides insight into the extent of the reaction; a high \( K_c \) value suggests that the reaction strongly favors the formation of products.
Knowing \( K_c \) allows us to judge whether a reaction mixture is at equilibrium or tell in which direction it needs to shift by comparing it with \( Q_c \). When \( Q_c < K_c \), more product formation is needed to reach equilibrium.
Direction of Chemical Reactions
Understanding the direction of a chemical reaction is crucial when dealing with equilibrium. The comparison between the reaction quotient \( Q_c \) and the equilibrium constant \( K_c \) allows us to predict this direction.
In our example, since \( Q_c = 0.686 \) is less than \( K_c = 4.7 \), the reaction will proceed in the forward direction to form more products. This is because the system will always strive to reach a state where \( Q_c = K_c \).
If \( Q_c \) were greater than \( K_c \), the reaction would move in the reverse direction, creating more reactants until equilibrium is established.
Industrial Synthesis of Hydrogen
Industrial synthesis of hydrogen is a practical application of chemical equilibrium concepts. It typically involves the reaction of steam and methane to produce synthesis gas, which contains hydrogen and carbon monoxide.
This process is crucial in industries as hydrogen is a key component in many applications, such as refining and ammonia production. Understanding the conditions under which this reaction best proceeds—like temperature, pressure, and reactant concentrations—is essential for optimizing production.
Furthermore, knowing when the reaction reaches equilibrium through concepts like \( Q_c \) and \( K_c \) ensures the efficiency and success of the hydrogen production process.
Chemical Equilibrium Calculations
Chemical equilibrium calculations allow chemists to predict the concentrations of reactants and products at equilibrium. By knowing the initial concentrations and using \( K_c \), it is possible to determine the final equilibrium state.
In the example of hydrogen synthesis, given concentrations are plugged into the formula for \( Q_c \) to see how the reaction is proceeding. Calculating \( Q_c \) and comparing it to \( K_c \) informs us about the necessary shift in the reaction.
This type of calculation is essential for managing reactions in both laboratory and industrial settings, providing control over product yields and resource efficiency.

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

Does the number of moles of products increase, decrease, or remain the same when each of the following equilibria is subjected to an increase in pressure by decreasing the volume? (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) (b) \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{C}(s)+\mathrm{CO}_{2}(g)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\)

Identify the true statement about the rate of the forward and reverse reaction once a reaction has reached equilibrium. (a) The rate of the forward reaction and the reverse reaction is zero. (b) The rate of the forward reaction is greater than the rate of the reverse reaction. (c) The rate of the reverse reaction is greater than the rate of the forward reaction. (d) The rate of the forward reaction is equal to the rate of the reverse reaction.

Ethylene glycol, used as antifreeze in automobile radiators, is manufactured by the hydration of ethylene oxide. Write the equilibrium constant expression for \(K_{c}\) \(\mathrm{O}\) \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(\mathrm{soln})+\mathrm{H}_{2} \mathrm{O}(\mathrm{soln}) \rightleftharpoons \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{soln})\)

Consider the following gas-phase reaction: \(2 \mathrm{~A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) An equilibrium mixture of reactants and products is subjected to the following changes: (a) A decrease in volume (b) An increase in temperature (c) Addition of reactants (d) Addition of a catalyst (e) Addition of an inert gas Which of these changes affect the composition of the equilibrium mixture but leave the value of the equilibrium constant \(K_{c}\) unchanged? Which of the changes affect the value of \(K_{c}\) ? Which affect neither the composition of the equilibrium mixture nor \(K_{c} ?\)

The reaction $$ 2 \mathrm{PH}_{3}(g)+\mathrm{As}_{2}(g) \rightleftharpoons 2 \mathrm{AsH}_{3}(g)+\mathrm{P}_{2}(g) $$ has \(K_{\mathrm{p}}=2.9 \times 10^{-5}\) at \(873 \mathrm{~K}\). At the same temperature, what is \(K_{\mathrm{p}}\) for each of the following reactions? (a) \(2 \mathrm{AsH}_{3}(g)+\mathrm{P}_{2}(g) \rightleftharpoons 2 \mathrm{PH}_{3}(g)+\mathrm{As}_{2}(g)\) (b) \(6 \mathrm{PH}_{3}(g)+3 \mathrm{As}_{2}(g) \rightleftharpoons 3 \mathrm{P}_{2}(g)+6 \mathrm{AsH}_{3}(g)\) (c) \(2 \mathrm{P}_{2}(g)+4 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}_{2}(g)+4 \mathrm{PH}_{3}(g)\)

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.