Consider the reaction between iodine gas and chlorine gas to form iodine monochl

Question

Consider the reaction between iodine gas and chlorine gas to form iodine monochloride:
I2(g)+Cl2(g)2ICl(g)Kp=81.9 (at 298 K)
A reaction mixture at 298 K initially contains PI2=0.30 atm and PCl2=0.30 atm .

Part A

What is the partial pressure of iodine monochloride when the reaction reaches equilibrium?

Consider the reaction between iodine gas and chlorine gas to form iodine monochloride:
I2(g)+Cl2(g)2ICl(g)Kp=81.9 (at 298 K)
A reaction mixture at 298 K initially contains PI2=0.30 atm and PCl2=0.30 atm .

Part A

What is the partial pressure of iodine monochloride when the reaction reaches equilibrium?

Explanation / Answer

cosider the given reaction

I2 + Cl2 ---> 2 ICl

the expression for Kp is given by

Kp = (pICl)^2 /(pI2) (pCl2)

now

consider the given reaction

I2 + Cl2 ---> 2 ICl

using ICE table

initial pressure of I2, Cl2 , ICl are 0.3 , 0.3 , 0

change in pressure of I2 , CL2 , ICl are -x , -x , +2x

equilibrium pressure of I2 ,Cl2 ,ICl are 0.3-x ,0.3-x , 2x

now

Kp = (pICl)^2 / (pI2) (pCl2)

81.9 = (2x)^2 / (0.3-x) (0.3-x)

81.9 = [ 2x / (0.3-x)]^2

9.05 = 2x / (0.3-x)

2x = 9.05 (0.3-x)

x = 0.2457

now

at equilibrium

pICl = 2x = 2 * 0.2457 = 0.4914

so

the partial pressure of iodine monchloride at equilibrium is 0.4914 atm

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