Question

The equilibrium constant, K_c, for the following reaction is 1.80×10^-2 at 698 K.

2HI(g) -->H₂(g) + I₂(g)

Calculate the equilibrium concentrations of reactant and products when 0.257 moles of HI are introduced into a 1.00 L vessel at 698 K.

[HI]	=	M
[H2]	=	M
[I2]	=	M

The equilibrium constant, K_c, for the following reaction is 83.3 at 500 K.

PCl₃(g) + Cl₂(g) -->PCl₅(g)

Calculate the equilibrium concentrations of reactant and products when 0.420 moles of PCl₃ and 0.420 moles of Cl₂ are introduced into a 1.00 L vessel at 500 K.

[PCl3]	=	M
[Cl2]	=	M
[PCl5]	=	M

Answer 1

Consider reaction,2 HI (g) H ₂ (g) + I ₂ (g)

Equilibrium constant for above reaction is , K c= [ H ₂  ] [ I ₂ ] / [ HI ] ² = 0.0180

Let's use ICE table.

M	2 HI (g) H 2 (g) + I 2 (g)
I	0.257
C	- 2X	+X	+X
E	0.257 - 2 X	X	X

K c = (X) (X) / ( 0.257 - 2 X ) ² = 0.0180

X ² / ( 0.257 - 2 X ) ² = 0.0180

Taking square root on both sides, we get X / 0.257 - 2 X = 0.1342

X = 0.1342 ( 0.257 - 2 X)

X = 0.03449 - 0.2684 X

X + 0.2684 X = 0.03449

1.2684 X = 0.03449

X = 0.03449 / 1.2684

X= 0.0272 M = [ H ₂  ] = [ I ₂ ]

Equilibrium concentration of HI = 0.257 - 2 X = 0.257 - 2 ( 0.0272 ) = 0.203 M

[ H ₂  ] = 0.0272 M

[ I ₂ ] = 0.0272 M

[ HI ] = 0.203 M

Question 2

Consider reaction, PCl ₃ (g) + Cl ₂ (g)PCl ₅ (g)  

Equilibrium constant for above reaction is , K c= [ PCl ₅ ] / [PCl ₃ ] [Cl ₂ ] = 83.3

Let's use ICE table.

M	PCl 3 (g) + Cl 2 (g)PCl 5 (g)
I	0.420	0.420
C	-X	-X	+X
E	0.420 -X	0.420-X	X

K c = X / ( 0.420 - X ) ( 0.420 -X ) = 83.3

X / 0.1764 - 0.420 X - 0.420 X + X ² = 83.3

X / 0.1764 - 0.840 X + X ² = 83.3

X = 83.3 ( 0.1764 - 0.840 X + X ² )

X = 14.694 - 69.972 X + 83.3 X ²

83.3 X ² - 69.972 X - X + 14.694 = 0

83.3 X ² -70.972 X + 14.694 = 0

Comparing to a X ² + b X + c = 0 , we get a =83.3, b = - 70.972 and c = 14.694

X = -b +/- b ² - 4 a c / 2 a

X = - ( -70.972 )  +/- ( -70.972 ) 2 -4 (1)( 14.694 ) / 2 (83.3)

X = 0.4975 M and X = 0.3547 M

We choose X = 0.3547 M

Equilibrium concentration of PCl ₃ = 0.420 - X = 0.0653 M

Equilibrium concentration of Cl ₂ = 0.420 - X =0.0653 M

Equilibrium concentration of  PCl ₅ = X = 0.3547 M

ANSWER :

[ PCl ₃ ] = 0.0653 M

[ Cl ₂ ] = 0.0653 M

[ PCl ₅ ] = 0.355 M