**Equilibrium Partial Pressure Calculation****Scenario:**A reaction vessel contains gases A and B. Initially, the pressure of gas A is 0.50 atm, and the pressure of gas B is 0.590 atm. The reaction of interest is:\[ \text{A (g) } \rightleftharpoons 2 \text{ B (g)} \]**Problem:**Determine the equilibrium partial pressure of B. The equilibrium constant (Kp) for this reaction is 67.2.**Procedure:**1. **Initial Conditions:** - Pressure of A: 0.50 atm - Pressure of B: 0.590 atm2. **Calculator Interface:** - A numerical keypad is shown to compute the required values, with operations including addition, multiplication, and clear/reset functions.3. **Objective:** - Use the equilibrium expression for the reaction and given Kp to solve for the equilibrium partial pressures of A and B. \[K_p = \frac{{(P_B)^2}}{{P_A}}\]Given the initial and equilibrium data, set up the equation to find the changes in pressure and solve for the unknowns.

A reversible reaction is defined as equilibrium reaction, when the rate of forward reaction becomes equal to the rate of backward reaction. Given reaction: Ag⇋ 2 Bg Initial pressure of A= 0.50 atm Initial pressure of B = 0.590 atm Equilibrium constant Kp = 67.2 According to law of equilibrium: Ag⇋ 2 Bgintial 0.50 0.590change -x +2xeqlb. 0.50-x 0.590+2x Equilibrium constant of the reaction is given by: Kp = PB2PA Substitute, 67.2 equilibrium constant, 0.50-x equilibrium pressure of A and 0.590+2x equilibrium pressure of B in expression: 67.2=0.590+2x20.50-x33.6 -67.2x = 0.3481 +4x2 +2.36x4x2+69.56x-33.2519 =0 Solving the quadratic equation: a=4, b=69.56 and c=-33.2519x=-b±b2-4ac2aSubstitute in the expression:x=-69.56+69.562-4×4×-33.25192×4x≈0.47 x=-69.56-69.562-4×4×-33.25192×4x≈-17.86 Pressure can not be in negative Therefore, the change at equilibrium is x ≈0.47 Equilibrium pressure of B = 0.590 +x atm Substitute, 0.47 as x in the expression: Equilibrium pressure of B =1.06 atm Therefore, equilibrium pressure of B is 1.06 atm.

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A reaction vessel is charged with 0.50 atm of A and 0.590 atm of B. Once the reaction reaches equilibrium, what is the equilibrium partial pressure of B? Kp for this reaction is 67.2 A (g) = 2 B (g)

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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