The equilibrium constant, \( K_c \), for the following reaction is \( 7.00 \times 10^{-5} \) at 673 K.\[ \text{NH}_4\text{I}(s) \rightleftharpoons \text{NH}_3(g) + \text{HI}(g) \]If an equilibrium mixture of the three compounds in a 4.33 L container at 673 K contains 1.71 mol of \( \text{NH}_3 \) and 0.445 mol of \( \text{NH}_4\text{I}(s) \), the number of moles of \( \text{HI} \) present is \(\_\_\_\_\_\_\) mol.

Answered: The equilibrium constant, Ke, for the…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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The equilibrium constant, \( K_c \), for the following reaction is \( 7.00 \times 10^{-5} \) at 673 K.

\[ \text{NH}_4\text{I}(s) \rightleftharpoons \text{NH}_3(g) + \text{HI}(g) \]

If an equilibrium mixture of the three compounds in a 4.33 L container at 673 K contains 1.71 mol of \( \text{NH}_3 \) and 0.445 mol of \( \text{NH}_4\text{I}(s) \), the number of moles of \( \text{HI} \) present is \(\_\_\_\_\_\_\) mol.

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This problem is based on equilibrium. We know that for solids the concentration is taken as 1 . So by that assumption we can calculate the moles of HI at equilibrium. The solution is given below.

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