Question

Help with both questions please!

6. The following reaction takes place in a 1.00L vessel at 500°C 2 Hlig Hae late Equilibrium concentrations were found to be 1.76 mol/L HL 0.200 mol/L. Hy and 0.200 mol/Liina vessel. If an additional 0.500 mol of hydrogen iodide gas is introduced at the same temperature, what are the new concentrations of all gases once equilibrium has been reestablished? A) Use the given concentrations to calculate Ke B) After the addition of an 0.500 mol of Hip- 2 HIC - Hate + late K=1.30x102 (7. The dissociation of ammonia to nitrogen and hydrogen gases at 400.0°C has a Ke value of 1.92. If 0.500 mol of ammonia is placed in a 500.0 ml container, determine the equilibrium concentrations of all gases. (Use quadratic formula solver)

Answer 1

Question 6:

a)

Key = [Products] Reactants [H212] (0.2002 TH 712 = 71 762 = 1.29 X 10

b)

After the addition of HI, the reaction will again going in forward direction

Using the ICE table

	[HI]	[H2]	[I2]
Initial	1.76 + 0.500	0.200	0.200
Change	-2x	x	x
Final	2.26-2x	0.200+x	0.200+x

2 K = Products (0.200 + Reactants = 7962 = 1.30 x 10-2

I=0.047

H2 = 12 = 0.247

HI] = 2.26 – 20.047) = 2.26 - 0.094 = 2.166M

Question 7:

2NH3(g) = N2(g) + 3H2(9

Using the ICE table

	[NH3]	[N2]	[H2]
Initial	1	0	0
Change	-2x	x	3x
Final	1-2x	x	3x

K = [Products] [Reactants] [N2] [H23 [N H3 2.(3.c) (1-2.c)2

x = 0.31

[NH3] = 1 - 2(0.31) = 1 - 0.62 = 0.38

[N2] = 0.31

[H2] = 3(0.31) = 0.93

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