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1) The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) <---...

1) The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) <--- --->PCl3(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.279 moles of PCl5(g) are introduced into a 1.00 L vessel at 500 K. [PCl5] = M [PCl3] = M [Cl2] = M 2) The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K. H2 (g) + I2 (g) <---- ----> 2 HI (g) Calculate the equilibrium concentrations of reactants and product when 0.388 moles of H2 and 0.388 moles of I2 are introduced into a 1.00 L vessel at 698 K. [ H2 ] = M [ I2 ] = M [ HI ] = M

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Expert Solution

1) The given reaction is,

_______PCl₅(g) <--> PCl₃(g)______ +_______ Cl₂(g)

I ______0.279 _______0 ______________________0

C_________-x _______+x _____________________+x

E_______0.279 -x ______x _____________________x

Now,

K_c = [PCl₃][Cl₂]/[PCl_5] = x * x / (0.279 - x) = 1.20 x 10^-2

=> x² / 0.279 - x = 1.20 x 10^-2

=> x² = (0.279 -x) 1.20 x 10^-2

=> x² = 0.279 x 1.20 x 10^-2 - 1.20 x 10^-2 * x

=> x² +1.20 * 10^-2 * x - 0.279 *1.20 * 10^-2 = 0

=> x² + 0.0120 x - 0.003348 = 0

Using quadratic formula , taking only posiitve value of x we get,

x = -0.0120 + 0.1163 / 2

=> x = 0.05215

[PCl₃ ]= 0.05215 mol / 1 L = 0.05215 M

[Cl₂] = 0.05215 mol / 1 L = 0.05215 M

and [PCl₅] = 0.279 -0.05215 mol / 1 L = 0.22685 M

2) The given reaction is H₂ (g) + I₂ (g) <---- ----> 2 HI (g)

ICE table H₂ (g) ______+_______ I₂ (g) <---- ----> 2 HI (g)

I______0.388 ________________0.388 __________0

C_____-x _____________________-x____________+2x

E_____0.388 -x _______________0.388-x ________ 2x

Now, K_c = [HI]² / [H₂] [I₂] = 55.6

=> 55.6 = (2x)² / (0.388 -x )* (0.388 -x )

=> 55.6 = 4x² / (0.388 -x)²

=> 55.6 * (0.388 -x)² =4 x²

=> 55.6 *( 0.150544 - 0.776x + x²) = x²

=> 8.3702464 - 43.1456 x + 55.6x² -4x² = 0

=> 51.6 x² -43.1456 x + 8.3702464 = 0

Using quadratic formula, we get two values of x as--

x = 43.1456 ( +-) 11.5725 / 103.2

=> x = 43.1456 + 11.5725/ 103.2 = 0.530 , or x = 43.1456 -11.5725 / 103.2 = 0.306

With 0.0.530 we egt negative value of concentration, so it cannot be used, therefore the accepted value of x will be 0.306

Now, [H₂] = 0.388 - 0.306 mol / 1 L = 0.082 M

[I₂] = 0.388 - 0.306 mol / 1 L = 0.082 M

[HI] = 2 x 0.306 mol / 1 L = 0.612 M

queen_honey_blossom answered 1 year ago

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