### Calculating \( K_p \) from \( K_c \)The equilibrium constant, \( K_c \), for the following reaction is \( 5.10 \times 10^{-6} \) at 548 K. Calculate \( K_p \) for this reaction at this temperature.\[ \text{NH}_4\text{Cl}(s) \rightleftharpoons \text{NH}_3(g) + \text{HCl}(g) \]#### Provided Data:- \( K_c = 5.10 \times 10^{-6} \)- Temperature, \( T = 548 \) K#### Formula to convert \( K_c \) to \( K_p \):\[ K_p = K_c(RT)^{\Delta n} \]Where:- \( R = 0.0821 \ \text{L·atm·K}^{-1}\text{·mol}^{-1} \) (Ideal gas constant)- \( \Delta n \) is the change in the number of moles of gas (products - reactants)For the given reaction:\[ \Delta n = (1 + 1) - (0) = 2 \]### Calculation:1. Substitute the known values into the formula: \[ K_p = K_c \times (R \times T)^{\Delta n} \] \[ K_p = 5.10 \times 10^{-6} \times (0.0821 \times 548)^2 \]2. Calculate \( (0.0821 \times 548) \): \[ (0.0821 \times 548) = 44.991 \]3. Raise this result to the power of 2: \[ (44.991)^2 = 2024.91 \]4. Multiply by \( K_c \): \[ K_p = 5.10 \times 10^{-6} \times 2024.91 = 1.03 \times 10^{-2} \]### Final Answer:\[ K_p = 1.03 \times 10^{-2} \]

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Answered: The equilibrium constant, K., for the…

The equilibrium constant, K., for the following reaction is 5.10x10-6 at 548 K. Calculate K, for this reaction at this temperature. NHẠCI(s) 2NH3(g) + HCl(g) Kp- =

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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