h Which of the above changes results in changing the value ofK,? canalyst 14. mi

Question

h Which of the above changes results in changing the value ofK,? canalyst 14. mixture ofo7s mol of N and 120 mol are placed in a container with an iron 3.0L and heated under pressure to 750 a.when reaction reaches equilibrium. IHJ-0.10M. What is the value ofNal and NHuat b 30 mol of ammonia are removed through condensation and no other gases are added. the new INJ, and NHJ when the reaction re-establishes equil what are Ke-128x or 1.3 3. (HI) 0.0153 M 0.000328 M 32: b. 1.2x10 0,16 atm, P 0.34 atm 5, 0.68 mol: P 6 atm, P 6, 16, 35, high temp [Cll 0.016 M. e. [NOCI-1.0M. DNOl 0.0.0080 M ICI.) 0.25 M & K, 68 x 10 9. a. &5x 103 M: b. 3.0x10 a M; c. x 8.0 x 10 M ID) 2.0 M: d. 30 x 10. [Fe' 0.5193 M, [Hg, 1-0.5096 M.[Fe 1-0.0107M, [Hig 0.0107 M ll. IBr l"0.167 M. ICOBr l-0.583 M, [Coj 0.667 M 12, 0.154 13, a, no change, decrease, Pro increases, no change, increase, no change, increase, increases b. increasing temperature only 14, a IN,

Explanation / Answer

Initially

M = mol/V

[N2] = 0.75/3 =0.25

[H2] = 1.2/3 = 0.4

[NH3] = 0

in equilbirium

[N2] = 0.25 - x

[H2] = 0.4 - 3x

[NH3] = 0 +2x

and we know

[H2] = 0.4 - 3x = 0.1

so

x = (0.1-0.4 )/-3 = 0.1

[N2] = 0.25 -  0.1 = 0.15

[H2] = 0.4 - 3* 0.1 = 0.10

[NH3] = 0 +2* 0.1 = 0.20

b)

if we add

0.3 mol --> 0.3/3 = 0.1 M of NH3, find new equilibrium

We need K

K = [NH3]^2/([N2][H2]^3)

K = (0.2^2)/((0.15)(0.10^3)

K = 266.66

so

new conditions:

[N2] = 0.15 - x

[H2] = 0.10 -3x

[NH3] = (0.20+0.10) + 2x

substitute

266.66 = (0.3+2x)^2 / (0.15 - x)(0.10 -3x)^3

solve for x

x = 0.1

so

[N2] = 0.15 - x = 0.15-0.01 = 0.14

[H2] = 0.10 -3x = 0.10 - 3*0.01 = 0.07

[NH3] = (0.20+0.10) + 2x = 0.3+2*0.01 = 0.12

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