Question

At 25°C, gaseous decomposes to and to the extent that 10.3% of the original (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is 1.30 atm. Calculate the value of for this system. =

Answer 1

The question is incomplete, complete question is :

At 25°C, gaseous
`SO_2Cl_2` decomposes to
`SO_2(g)` and
`Cl_2(g)` to the extent that 10.3% of the original

Answer:

The value of the
`K_p` of the reaction is 0.0139.

Step-by-step explanation:

The total pressure at equilibrium = P = 1.30 atm.

`SO_2Cl_2(g)\rightleftharpoons SO_2(g) + Cl_2(g)`

Initially

1 mol 0 0

At equilibrium:

(1 - 0.103)mol 0.103 mol 0.103 mol

Total moles at equilibrium,n = (1-0.103)mol+ 0.103 mol + 0.103 mol

n = 1.103 moles

At equilibrium ,mole fraction of
`SO_2Cl_2=\chi_1=((1-0.103) mol)/(1.103mol)=0.8132`

Partial pressure of f
`SO_2Cl_2` equilibrium ;

`p_1=P* \chi_1` (Dalton's law of partial pressure)

`p_1=1.30* 0.8132=1.057 atm`

At equilibrium ,mole fraction of
`SO_2=\chi_2=((0.103) mol)/(1.103mol)=0.09338`

Partial pressure of f
`SO_2` equilibrium ;

`p_2=P* \chi_2` (Dalton's law of partial pressure)

`p_2=1.30* 0.09338=0.1214 atm`

At equilibrium ,mole fraction of
`Cl_2=\chi_3=((0.103) mol)/(1.103mol)=0.09338`

Partial pressure of f
`Cl_2` equilibrium ;

`p_3=P* \chi_3` (Dalton's law of partial pressure)

`p_3=1.30* 0.09338=0.1214 atm`

The value of the
`K_p` of the reaction will be;

`K_p=(p_2* p_3)/(p_1)=(0.1214 atm* 0.1214 atm)/(1.057 atm)=0.0139`