Question

B. The reaction A(g) + 2B(g) 2C(g) was allowed to come to equilibrium. The initial After the reaction reached equilibrium, the amount of B remaining in the 2.0 L vessel was 1.0 amounts of reactants placed into a 2.00 L vessel were 2,0 mol A and 2.0 mol B and 1,0 mol C. mol of B. Calculate (show formulas and appropriate steps in each case): a) (10 points) Ke for this reaction: Ec = LA)632 (2)(2) = to equilibrium for the process described above starting with the initial amounts stated. Label the axes. Label the curves. b) (20 points) Show a reasonable plot for the concentrations of A, B, and C during the approach

Answer 1

Initial amount of A = 2 moles

Initial amount of B = 2 moles

Initial amount of C = 1 moles

Volume of vessel = 2 liters

Initial concentration of A = number of moles of A/Volume of vessel = 2/2 = 1 M

Initial concentration of B = 2/2 = 1M

Initial concentration of C = 1/2 = 0.5 M

At equilibrium, amount of B = 1 mole

Equilibrium concentration of B = 1/2 = 0.5 M

The reaction is

A(g) + 2B(g) -------------> 2C(g)

Initial concentrations 1 1 0.5

Let x M of A be consumed at equilibrium

Equilibrium concentration 1-x 1-2x 0.5+2x

Given

[B]_equilibrium = 1-2x = 0.5 M

=> 2x = 0.5

=> x = 0.25

Therefore

equilibrium concentration of A = 1-0.25 = 0.75 M

Equilibrium concentration of C = 0.5+0.5 = 1 M

Therefore the equilibrum constant, K_c is

K_c = [C]²/([A][B]²]

= 1/(0.75*(0.5)²)

Equilibrium constant, K_c = 5.333

2)

Based on the information given above, the variation of the concentration of A, B and C with extent of reaction, x is given below. At x = 0.25, equilibrium is attained, so not further net reaction occurs

x	A = 1-x	B = 1-2x	C = 0.5+2x
0	1	1	0.5
0.05	0.95	0.9	0.6
0.1	0.9	0.8	0.7
0.15	0.85	0.7	0.8
0.2	0.8	0.6	0.9
0.25	0.75	0.5	1

This data has been plotted below

Concentration, M [A] +[B] +[C] 0.05 0.1 0.15 0.2 0.25 0.3 Extent fo reaction, x