Suppose a reaction has the equilibrium constant \(K=1.7 \times 10^{-8}\) at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

Short Answer

Expert verified
The equilibrium constant K for the reaction is given as \(1.7 \times 10^{-8}\), which is an extremely small value. This indicates that there will be a large amount of unreacted starting material and a small amount of products at equilibrium. Therefore, this reaction is not likely to be a good source of products at this temperature.

Step by step solution

01

Interpret the given equilibrium constant value

The equilibrium constant K is given as \(1.7 \times 10^{-8}\). Since this value is extremely small, it indicates that there will be a considerably larger amount of unreacted starting material (reactants) than products when the reaction reaches equilibrium.
02

Determine the amount of unreacted starting material at equilibrium

As we have concluded that the equilibrium constant K indicates a much higher concentration of reactants compared to products when the reaction is at equilibrium, it can be inferred that there will be a large amount of unreacted starting material present at equilibrium.
03

Evaluate the reaction as a source of products

Since the value of K is very small, resulting in a large amount of unreacted starting material and small amount of products at equilibrium, this reaction is not likely to be a good source of products at this temperature. Increasing the temperature or adding a catalyst may improve product formation in some cases.

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

Consider the reaction $$ \mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q) $$ How will the equilibrium position shift if a. water is added, doubling the volume? b. \(\mathrm{AgNO}_{3}(a q)\) is added? (AgSCN is insoluble.) c. \(\mathrm{NaOH}(a q)\) is added? \(\left[\mathrm{Fe}(\mathrm{OH})_{3}\right.\) is insoluble.] d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added?

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ If \(2.0 \mathrm{~mol} \mathrm{NO}\) and \(1.0 \mathrm{~mol} \mathrm{Cl}_{2}\) are placed into a \(1.0\) - \(\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\ \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\ \mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\ \mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14} \end{array} $$ determine the values for the equilibrium constants for the following reactions. a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

At a particular temperature, \(8.0 \mathrm{~mol} \mathrm{NO}_{2}\) is placed into a \(1.0\) -L container and the \(\mathrm{NO}_{2}\) dissociates by the reaction $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ At equilibrium the concentration of \(\mathrm{NO}(g)\) is \(2.0 \mathrm{M}\). Calculate \(K\) for this reaction.

A 1.00-L flask was filled with \(2.00\) mol gaseous \(\mathrm{SO}_{2}\) and \(2.00 \mathrm{~mol}\) gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that \(1.30\) mol gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K\), for this reaction.

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