For the reaction PCl 5 ( g ) ⇌ PCl 3 ( g ) + Cl 2 ( g ) at 600. K, the equilibrium constant, K p , is 11.5. Suppose that 2.450 g PCl 5 is placed in an evacuated 500.-mL bulb, which is then heated to 600. K. a . What would be the pressure of PCl 5 if it did not dissociate? b . What is the partial pressure of PCl 5 at equilibrium? c . What is the total pressure in the bulb at equilibrium? d . What is the percent dissociation of PCl 5 at equilibrium?

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of PCl5=208.23 g⋅mol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The pressure of PCl5 in undissociated form.

Explanation

The pressure of PCl5 in undissociated form is 1.16 atm_ .

Given

The reactions are given as,

PCl5(s)⇌PCl5(g)+Cl2(g)

The equilibrium constant Kp=11.5 at 600 K .

Mass of PCl5=2.450 g .

Volume of the PCl5 =500 ml .

Moles of the PCl5 is determined by the formula,

n=mM

Where,

• n is the number of moles.
• m is the given mass.
• M is the molar mass.

Substitute the values in the above equation.

n=mM=2.450 g208

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of PCl5=208.23 g⋅mol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The partial pressure of PCl5 at equilibrium.

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of PCl5=208.23 g⋅mol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The total pressure in the bulb at equilibrium.

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of PCl5=208.23 g⋅mol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The percent dissociation of PCl5 at equilibrium.