Can someone please help solve the steps taken to answer the 2 questions below. T

Question

Can someone please help solve the steps taken to answer the 2 questions below. The answer for question 1 is 3.09, and the answer for question 2 is 1.2%

1 atm 1.01325 bar, 1 bar 102 Pa, Pa s arms 1.01323 din bp, vent X solvent X po i m i ma Kfp, ATbp CRT, AT Question 1 (10 points): The chemical reaction A(g) 3 B(g) is carried out in a rigid container at a constant temperature of 17.9 oC. At the start of the reaction 2.00 atm of A and 7.00 atm ofc are added to the container, total pressure of 9.00 atm. At the end of the reaction, the total pressure in the container is 11.00 atm. Calculate the equilibrium constant of the reaction. Question 2 (10 points): At 25 oC the equilibrium constant of the acid HA is 6.3 x 10 6. Calculate the percent ionization lyield) of a 0.045 M solution of the acid. ionization weak acid Page

Explanation / Answer

Q1.

A + 3B = 2C

initially

P-A = 2

P-B = 0

P-C =7

in equilibrium

P-A = 2 + x

P-B = 0 + 3x

P-C =7 - 2x

final P = 11

Pa + Pb + Pc = 11

2 + x +  0 + 3x + 7 - 2x = 11

9 + 4x-2x = 11

2x = 2

x = 1

P-A = 2 + x = 2+1 = 3

P-B = 0 + 3x = 3*1 = 3

P-C =7 - 2x = 7-2 = 5

claculate Kp

Kp = P-C^2 /(PA)(PC^3)

Kp = (5^2)/((3)(3^3))

Kp = 0.308641

Q2.

Ka = 6.3*10^-6

Ka = [H+][A-]/[HA]

6.3*10^-6 = x*x/(0.045-x)

x = H = 5.29*10^-4

% ion = [H+]/M*100% = (5.29*10^-4/(0.045) * 100 = 1.17555 % = 1.2%

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---