The equilibrium partial pressures of Br 2 , Cl 2 , and BrCl will be the same as

Question

The equilibrium partial pressures of Br2, Cl2, and BrCl will be the same as the initial values.

The equilibrium partial pressure of Br2 will be greater than 1.00 atm.

At equilibrium, the total pressure in the vessel will be less than the initial total pressure.

The equilibrium partial pressure of BrCl (g) will be greater than 2.00 atm.

The reaction will go to completion since there are equal amounts of Br2 and Cl2.

The equilibrium partial pressures of Br2, Cl2, and BrCl will be the same as the initial values.

The equilibrium partial pressure of Br2 will be greater than 1.00 atm.

At equilibrium, the total pressure in the vessel will be less than the initial total pressure.

The equilibrium partial pressure of BrCl (g) will be greater than 2.00 atm.

The reaction will go to completion since there are equal amounts of Br2 and Cl2.

Explanation / Answer

For the case in which:

Br2 + Cl2 = 2BrCl

then

Kp = P-BrCl^2 /(P-Br2)(P-Cl2)

so

Q =  P-BrCl^2 /(P-Br2)(P-Cl2)

Substitute

Q = (2^2)/(1)(1) = 4

since Q < K

This means that initially, there is more BrCl than Br Cl than the equilibrium allows, meaning that eventually

BrCl decomposes to form more Br2 and Cl2

so

This is true ---> The equilibrium partial pressure of Br2 will be greater than 1.00 atm, since it will increase

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