Question

# For the equilibrium Br2(g) + Cl2(g) ⇌ 2 BrCl(g) the equilibrium constant Kp is 7.0 at...

For the equilibrium Br2(g) + Cl2(g) ⇌ 2 BrCl(g) the equilibrium constant Kp is 7.0 at 400 K. If a cylinder is charged with BrCl(g) at an initial pressure of 1.00 atm and the system is allowed to come to equilibrium what is the final (equilibrium) pressure of BrCl?

For the equilibrium  the equilibrium constant  is 7.0 at 400 . If a cylinder is charged with  at an initial pressure of 1.00  and the system is allowed to come to equilibrium what is the final (equilibrium) pressure of ?

 0.31 atm 0.15 atm 0.45 atm 0.57 atm 0.22 atm

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually H2SO4). A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water:
CH3COOH(solv)+CH3CH2OH(solv)⇌CH3COOCH2CH3(solv)+H2O(solv)
where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at 55 ∘C is 6.68. A pharmaceutical chemist makes up 15.0 L of a solution that is initially 0.275M in acetic acid and 3.85M in ethanol.

At equilibrium, how many grams of ethyl acetate are formed?

Kp=p(Br2)*p(CL2)/p(BrCl)^2

ICE table

 p[BrCl] p[Cl2] p[Br2] Initial 1.00 atm 0 0 change -2x +x +x equilibrium 1.00-2x x x

Kp=7.0=x^2/(1.00-2x)^2

7.0=x^2/1+4x^2-4x

7+28x^2-28x=x^2

26x^2-28x+7=0

X=0.39atm=p(cl2)=pBr2

Equilibrium pressure of BrCl=1.00-2*0.39=0.22 atm

CH3COOH(solv)+CH3CH2OH(solv)⇌CH3COOCH2CH3(solv)+H2O(solv)

Kc=6.68=[ CH3COOCH2CH3][H2O]/[ CH3COOH][ CH3CH2OH]

ICE table

 [CH3COOH] [CH3CH2OH] [CH3COOCH2CH3] [H2O] Initial 0.275 M 3.85M 0 0 change -x -x +x +x equilibrium 0.275-x 3.85-x x x

Kc=6.68=x^2/(0.275-x)(3.85-x)

6.68=x^2/(0.275-x)(3.85-x)

Or,7.68x^2+26.89x-7.09=0

Solving for x,

X=0.25M=[CH3COOCH2CH3]=eqm conc

Volume=15.0L

Eqm mass of ethyl acetate=0.25 M*15.0L=0.25 mol/L*15L=3.75 g

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