The equilibrium constant, \( K_c \), for the following reaction is \( 5.10 \times 10^{-6} \) at 548 K.\[\text{NH}_4\text{Cl}(g) \rightleftharpoons \text{NH}_3(g) + \text{HCl}(g)\]Calculate the equilibrium concentration of HCl when 0.230 moles of \(\text{NH}_4\text{Cl}(s)\) are introduced into a 1.00 L vessel at 548 K.\[[\text{HCl}] = \underline{\hspace{2cm}}\text{M}\]Buttons:- **Submit Answer**- **Retry Entire Group**A note indicates that there are 9 more group attempts remaining. Navigation:- **Previous**- **Next**There is a hint available, denoted by "Show Hint."[This information is provided on a learning platform interface.]

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 \( 5.10 \times 10^{-6} \) at 548 K.

\[
\text{NH}_4\text{Cl}(g) \rightleftharpoons \text{NH}_3(g) + \text{HCl}(g)
\]

Calculate the equilibrium concentration of HCl when 0.230 moles of \(\text{NH}_4\text{Cl}(s)\) are introduced into a 1.00 L vessel at 548 K.

\[
[\text{HCl}] = \underline{\hspace{2cm}}\text{M}
\]

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Step 1

K_c is the ratio of product of concentration of products raised to their stoichiometric coefficient to the product of concentration of reactants raised to their stoichiometric coefficients.

Molarity is the ratio of number of moles to the volume of solution in liters

Note :- The solution of this problem is in accordance with considering the physical state of NH₄Cl as gas.

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