im022, 9:20 amHW4, Ch14Q1.Hint:InitialChangeEquil32:56/59-23F5H₂(g) + [₂(g) = 2 HI(g), K₂ = 54.3 at 430°CInital CH2=0.40M0.40OOK₂=54.3 =13ZxFOxx(0.40-22)²(*) (*)MacBook AirF7▶11-2x0.40-2XandSolve for xfind equilibriumcone of each Species

H2(g) + I2(g) <------> 2HI(g) Kc = 54.3 Initial concentration of HI = 0.40 M

9:20 am HW4, Ch14 Q1. at: Initial Сваде Equil H₂(g) + [2(g) = 2 HI(g), K₂ = 54.3 at 430°C 0.40 Inital CH2=0.40M 32-58/59-23 O N 2 x K₂=54.3 = O * א x (0.40-22)² (*) (*) <-2x 0.40-2X and Solve for x find equilibrium cone of each species

9:20 am HW4, Ch14 Q1. at: Initial Сваде Equil H₂(g) + [2(g) = 2 HI(g), K₂ = 54.3 at 430°C 0.40 Inital CH2=0.40M 32-58/59-23 O N 2 x K₂=54.3 = O * א x (0.40-22)² () () <-2x 0.40-2X and Solve for x find equilibrium cone of each species

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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Question

im
022, 9:20 am
HW4, Ch14
Q1.
Hint:
Initial
Change
Equil
32:56/59-23
F5
H₂(g) + [₂(g) = 2 HI(g), K₂ = 54.3 at 430°C
Inital CH2=0.40M
0.40
O
O
K₂=54.3 =
13
Z
x
FO
x
x
(0.40-22)²
(*) (*)
MacBook Air
F7
▶11
-2x
0.40-2X
and
Solve for x
find equilibrium
cone of each Species

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