**Equilibrium Reaction and Calculation of \( K_p \):**Gaseous \(\text{NH}_3\) is placed in a closed container at **33 °C**, where it partially decomposes to \(\text{N}_2\) and \(\text{H}_2\), according to the following equilibrium reaction:\[ 2 \text{NH}_3(g) \rightleftharpoons 1 \text{N}_2(g) + 3 \text{H}_2(g) \]At equilibrium, the partial pressures are measured as follows:- \( p(\text{NH}_3) = 0.008450\) atm,- \( p(\text{N}_2) = 0.007140\) atm,- \( p(\text{H}_2) = 0.003440\) atm.**Objective:**Determine the value of \( K_p \) for this reaction at 33 °C.**Calculation:**The equilibrium constant \( K_p \) for the reaction is expressed as:\[K_p = \frac{(p(\text{N}_2) \times (p(\text{H}_2))^3)}{(p(\text{NH}_3))^2}\]Insert the given partial pressures into the expression to find \( K_p \).\[ K_p = \frac{(0.007140) \times (0.003440)^3}{(0.008450)^2} \]**Note to Educators:**Ensure students are aware that the units for partial pressures must be consistent when calculating \( K_p \). This exercise is useful for reinforcing concepts of chemical equilibrium and the use of equilibrium constants in gaseous reactions.

Gaseous NH3 is placed in a closed container at 33 °C, where it partially decomposes to N₂ and H₂: 2 NH3(g) = 1 N₂(g) + 3 H₂(g) At equilibrium it is found that p(NH3) = 0.008450 atm, p(N₂) = 0.007140 atm, and p(H₂) = 0.003440 atm. What is the value of Kp at this temperature? Kp =

Gaseous NH3 is placed in a closed container at 33 °C, where it partially decomposes to N₂ and H₂: 2 NH3(g) = 1 N₂(g) + 3 H₂(g) At equilibrium it is found that p(NH3) = 0.008450 atm, p(N₂) = 0.007140 atm, and p(H₂) = 0.003440 atm. What is the value of Kp at this temperature? 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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