Question

Part A.) Phosgene (carbonyl chloride), COCl2, is an extremely toxic gas that is used in manufacturing certain dyes and plastics. Phosgene can be produced by reacting carbon monoxide and chlorine gas at high temperatures: CO(g)+Cl2(g)⇌COCl2(g) Carbon monoxide and chlorine gas are allowed to react in a sealed vessel at 475 ∘C . At equilibrium, the concentrations were measured and the following results obtained: Gas Partial Pressure (atm) CO 0.840 Cl2 1.28 COCl2 0.130. What is the equilibrium constant, Kp, of this reaction?

Part B.) The following reaction was performed in a sealed vessel at 797 ∘C :

H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.00M and [I2]=2.35M. The equilibrium concentration of I2 is 0.0700 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Answer 1

The expression for equilibrium constant ( K_p  ) concerning the given reaction -

CO(g) + Cl₂ (g) <---------------> COCl₂ (g) ---, is written as -

K_p   = P_COCl2   / P_CO(g)   x P_(Cl2(g)  , where P represents the partial pressures of respective gases at equilibrium state

Since equilibrium state partial pressures are given as

P_(COCl2(g)   = 0.130 atm : P_CO(g) = 0.84 atm , P_Cl2(g)   = 1.28

Substituting the given values in equillibrium state expression we get ,

K_p   = 0.130 atm / ( 0.840 atm x 1.28 atm)   

---- = 0.1209 atm^-1   

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Part B )

Similarly The Equilibrium state expression Kc for the reaction-

H2(g) + I2 (g) <------------> 2HI(g)

is Kc = [HI(g)]^2 / [H2(g)] [I2(g)]

where the equilibrium concentrations of different reactants and products are written within [.....]

An stoichiometric study of the given reaction reveals that the equilibrium concentrations of different reactants and products are -

[ I2 (g) ] = 0.0700 M

[ H2(g) ] = 3.00 - moles of H2 used for reaction ]

------------ = { 3.00 - (2.35- 0.0700) } , { since equqal number of moles of I2 & H2 are used for the reaction & no. of moles of i2 used is equal to (2.35 - 0.0700 )

------------- = 0.72 M

[ HI(g) ] = 2 x 2.28 , { because each mole of reactants H2 or I2 (any one in this case ) forms twice the --------------------------------------------------------number of moles of reactants used }

------------- = 4.56

Substituting the above values of equilbrium concentrations in the given expressions for Kc we get

Kc = (4.56)^2 / 0.07 x 0.72

---- = 412.57

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