Question

1. The equilibrium constant, K_p, for the following reaction is 10.5 at 350 K:

2CH₂Cl₂(g) <--->CH₄(g) + CCl₄(g)

Calculate the equilibrium partial pressures of all species when CH₂Cl₂(g) is introduced into an evacuated flask at a pressure of 0.865 atm at 350 K.

PCH2Cl2	=____	atm
PCH4	=____	atm
PCCl4	=____	atm

2. The equilibrium constant, K_p, for the following reaction is 0.215 at 673 K:

NH₄I(s) <----> NH₃(g) + HI(g)

Calculate the equilibrium partial pressure of HI when 0.413 moles of NH₄I(s) is introduced into a 1.00 L vessel at 673 K.

P_HI = ____atm

Answer 1

2CH2CL2(g) --------> CH4(g) + CCl4(g)

0.865 atm 0 0 ---------> initial

0.865-2x x x ---------> equilibrium

K_P = (P_CH4 P_CCl4) / (P_CH2Cl2)² = 10.5

= x .x / (0.865-2x)² = 10.5

= (x / (0.865-2x))² = 10.5

= x / (0.865-2x) = √10.5

= x / (0.865-2x) = 3.24

x = 3.24 (0.865-2x)

x = 2.8026 - 6.48x

=> 7.48 x = 2.8026

=> x = 2.8026 / 7.48

=> x = 0.375

P_CH4 = P_CCl4 = x = 0.375 atm

P_CH2Cl2 = (0.865 - 2x) = (0.865 - 2(0.375)) = 0.115 atm

P_CH2Cl2 = 0.115 atm

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K_P = 0.215 , T = 673K V= 1L

NH4I <----> NH3 + HI

n= 0.413 moles

consider ideal gas equation

PV = nRT

P = nRT / V

= (0.413mol x 0.08206 L atm mol^-1 K^-1 x 673 K) / 1 L

P = 22.808 atm

NH4I <---->   NH3 + HI

22.808 atm 0 0 ---------> initial

22.808-x x x ---------> equilibrium

K_P = (P_NH3 P_HI) / (P_NH4I) = 0.215

= (x .x) / (22.808-x) = 0.215

= (x² / (22.808-x)) = 0.215

= x² = 0.215(22.808-x)

=> x² = 4.904 - 0.215x

=> x² + 0.215x - 4.904 = 0

=> x = 2.1096

therefore the partial pressure of HI is 2.1096 atm