McGraw-Hill Connect-E Homework #2 Previous Attempt View xC Search Q&A; x x -) De
ID: 935238 • Letter: M
Question
McGraw-Hill Connect-E Homework #2 Previous Attempt View xC Search Q&A; x x -) Dezto.mheducation.com/hm.tpx E connect General Chemistry CH 22 CRN 31193 11 HEMISTRY Homework #2 prev Question #9 (of 14) | next 9· 7.14 points Question Be sure to answer all parts. As The dissociation of molecular iodine into iodine atoms is represented as 12(g) 21(g) At 1000 K, the equilibrium constant Kc for the reaction is 3-80-10-5. Suppose you start with 0.0451 mol of I2 in a 2.35-L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of I2? What is the equilibrium concentration of I?Explanation / Answer
I2 <-> 2I
Kc = 3.8*10^-5
0.0451 mol of I2
V =2.35 L
Concentration of gases in equilibrium?
Kc = [I]2 / [I2]
Initial Concentration
I = 0 M
I2 = 0.0451 mol / 2.35L = 0.01919 M
Final concnetration
I = +2x
I2 = 0.01919 - x
Substitutein Kc
Kc = [I]2 / [I2]
3.8*10^-5 = (2x)^2 / (0.01919-x)
Solve for x
3.8*10^-5 = 4x^2 / (0.01919-x)
(0.01919-x)(3.8*10^-5) = 4x^2
7.29*10^-7 -(3.8*10^-5) x -4x^2 = 0
x = 4.22*10^-4
Substitute in concentrations:
I = +2x = 2(4.22*10^-4) = 8.44*10^-4
I2 = 0.01919 - x = 0.01919 - 4.22*10^-4 = 0.01876
[I] = 8.44*10^-4
[I2] = 0.01876