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

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant, (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst?

Short Answer

Expert verified
The equilibrium constant for a gas-phase exothermic reaction remains unchanged upon the removal of a reactant (a) or product (b), as well as upon a decrease in volume (c) and the addition of a catalyst (e), due to its dependence solely on temperature. However, when the temperature decreases (d), the equilibrium constant will also decrease since the equilibrium shifts to the right, favoring the exothermic forward reaction.

Step by step solution

01

(a) Removal of a reactant

Removing a reactant from the system will decrease its concentration. According to Le Chatelier's principle, when the concentration of a reactant decreases, the equilibrium will shift to the left to compensate for the change and form more reactants. However, this shift will not affect the value of the equilibrium constant, as it only depends on the temperature and not on the concentrations of the reactants or products.
02

(b) Removal of a product

Removing a product from the system will decrease its concentration. According to the Le Chatelier's principle, when the concentration of a product decreases, the equilibrium will shift to the right to compensate for the change and form more products. Again, this shift will not affect the value of the equilibrium constant, as it only depends on temperature and not on the concentrations of the reactants or products.
03

(c) Decrease in volume

A decrease in volume will result in an increase in pressure, assuming the temperature and the number of moles of gases remain constant. According to Le Chatelier's principle, when the pressure increases, the equilibrium will shift in the direction with fewer gaseous moles to minimize the change. However, this shift in equilibrium will not affect the equilibrium constant because, as mentioned earlier, the equilibrium constant depends only on temperature.
04

(d) Decrease in temperature

Since the reaction is exothermic, a decrease in temperature will favor the forward reaction because it releases heat. According to Le Chatelier's principle, the equilibrium will shift to the right, and more products will be formed to compensate for the decrease in temperature. This change in temperature will affect the value of the equilibrium constant, as it depends on temperature. In this case, the value of the equilibrium constant will decrease as the temperature decreases.
05

(e) Addition of a catalyst

The addition of a catalyst will increase the rate of both the forward and reverse reactions, resulting in a faster approach to equilibrium. However, a catalyst will not affect the value of the equilibrium constant because, as mentioned earlier, the equilibrium constant depends only on temperature and not on the rate at which 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 Vaia!

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

At \(800 \mathrm{~K}\), the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) is \(K_{c}=3.1 \times 10^{-5}\). If an equilibrium mixture in a 10.0-L vessel contains \(2.67 \times 10^{-2} \mathrm{~g}\) of \(\mathrm{I}(\mathrm{g})\), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

The equilibrium constant for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{NOBr}(g) $$ is \(K_{c}=1.3 \times 10^{-2}\) at \(1000 \mathrm{~K}\). (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor \(\mathrm{NOBr}\) ? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\). (c) Calculate \(K_{c}\) for \(\operatorname{NOBr}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\).

At \(2000^{\circ} \mathrm{C}\), the equilibrium constant for the reaction $$ 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) $$ is \(K_{c}=2.4 \times 10^{3}\). If the initial concentration of \(\mathrm{NO}\) is \(0.175 \mathrm{M}\), what are the equilibrium concentrations of \(\mathrm{NO}\), \(\mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) ?

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}(g) \rightleftharpoons \mathrm{Ni}(s)+4 \mathrm{CO}(g)\) (d) \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}(a q)\) (e) \(2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(a q) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Zn}(s)\) (f) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (g) \(2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+2 \mathrm{OH}(a q)\)

Consider the hypothetical reaction \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{C}(g)\), for which \(K_{c}=0.25\) at a certain temperature. A \(1.00-\mathrm{L}\) reaction vessel is loaded with \(1.00 \mathrm{~mol}\) of compound \(\mathrm{C}\), which is allowed to reach equilibrium. Let the variable \(x\) represent the number of \(\mathrm{mol} / \mathrm{L}\) of compound \(\mathrm{A}\) present at equilibrium. (a) In terms of \(x\), what are the equilibrium concentrations of compounds \(B\) and \(C\) ? (b) What limits must be placed on the value of \(x\) so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of \(x\) ) into the equilibriumconstant expression, derive an equation that can be solved for \(x\). (d) The equation from part (c) is a cubic equation (one that has the form \(a x^{3}+b x^{2}+c x+d=0\) ). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of \(x\) that you specified in part (b). The point at which the cubic equation crosses the \(x\)-axis is the solution. (e) From the plot in part (d), estimate the equilibrium concentrations of \(A, B\), and \(C\). (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

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