show work 10. For the reaction 2 A (g)+3 B (g-C (g)·2D (9), complete the followi

Question

show work 10. For the reaction 2 A (g)+3 B (g-C (g)·2D (9), complete the following table and find the value of the equilibrium constant (K) initial pressure (atm) 040 0.50 change due to reaction (D) in atm equilibrium pressure (atm) 0.26 11. For the reaction C (g) . 3 D (g)-A (g)-28 (g), complete the following table and find the value of the equilibrium constant (K.) 3D A2B 0.50 | 0.60 1.00 | 1.00 initial conc (M) change due to reaction (0) in M equilibrium conc (M) 0.80 12. Find the yalue of the equilibrium constant for No. (9)-. 2 No, (g) given initial concentrations: N° 1.00 atm; [No,1-0; the equilibrium conc. of NO2 is 040 atm. Page 8 of 9

Explanation / Answer

2A        +        3B        -->       C +       2D

Initial   0.4                   0.5                0             0

Change 2x                    3x            x               2x

Equilb    0.4-2x             0.5-3x           x              2x

As given the equilibrium pressure of B = 0.26 = 0.5-3x

So 3x = 0.24

x = 0.08 atm

So equilibrium pressures will be

A = 0.4-2x = 0.4 – 0.16 = 0.24atm

B = 0.5 – 3x = 0.5-3X0.08 = 0.26 atm

C = 0.08 atm

D = 2x = 0.16

Kp = P2A X P3B / PC X P2D = 0.24 X 0.24 X 0.26 X 0.26 X 0.26 / 0.08 X 0.16 X 0.16 = 0.494

A

B

C

D

Initial Pressure

0.4

0.5

0

0

Change

0.16

0.24

.08

.16

Equilibrium Pressure

0.24

0.26

.08

.16

A

B

C

D

Initial Pressure

0.4

0.5

0

0

Change

0.16

0.24

.08

.16

Equilibrium Pressure

0.24

0.26

.08

.16

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