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In: Chemistry

The equilibrium constant, K, of a reaction at a particular temperature is determined by the concentrations or pressures of the reactants and products at equilibrium.

 

The equilibrium constant, K, of a reaction at a particular temperature is determined by the concentrations or pressures of the reactants and products at equilibrium.

For a gaseous reaction with the general form aA+bB⇌cC+dD

the Kc and Kp expressions are given by Kc=[C]c[D]d/[A]a[B]b Kp=(PC)c(PD)d(PA)a(PB)b

The subscript c or p indicates whether K is expressed in terms of concentrations or pressures. Equilibrium-constant expressions do not include a term for any pure solids or liquids that may be involved since their composition does not change throughout the reaction. The standard state of a pure substance is the pure substance itself, and although the quantity may change the sample remain pure. The concentration is effectively equal to 1, and will not impact the magnitude of K.

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 479 ∘C . At equilibrium, the concentrations were measured and the following results obtained

Gas Partial Pressure
(atm )
CO 0.750
Cl2 1.23
COCl2 0.280

What is the equilibrium constant, Kp , of this reaction? Express your answer numerically.

PART B: The following reaction was performed in a sealed vessel at 729 ∘C :

H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.60M and [I2]=2.25M . The equilibrium concentration of I2 is 0.0500 M . What is the equilibrium constant, Kc , for the reaction at this temperature? Express your answer numerically.

Solutions

Expert Solution

CO(g) + Cl2(g) COCl2(g)

At equilibirium,

PCO = 0.75 atm, PCl2 = 1.23 atm and PCOCl2 =0.28 atm.

So, kp = (PCOCl2) / (PCO x PCl2)

= (0.28) / (0.75 x 1.23)

= 3.03 x 10-1

PART B:

H2(g) + I2(g) 2HI(g)

Initail concentration of [H2] = 3.6 M and [I2] = 2.25 M.

At equlibirium, [I2] = 0.05 M,

So. [H2] = 0.05 M and [HI] = (2 x 0.05) M = 0.1 M

So, Kc = [HI]2 / ( [H2] x [I2]) = (0.1)2 / (0.05 x 0.05) = 0.01/0.0025 =4

queen_honey_blossom answered 2 years ago
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