Respuesta :
The question is incomplete, complete question is :
At 25°C, gaseous [tex]SO_2Cl_2[/tex] decomposes to [tex]SO_2(g)[/tex] and [tex]Cl_2(g)[/tex] to the extent that 10.3% of the original
Answer:
The value of the [tex]K_p[/tex] of the reaction is 0.0139.
Explanation:
The total pressure at equilibrium = P = 1.30 atm.
[tex]SO_2Cl_2(g)\rightleftharpoons SO_2(g) + Cl_2(g)[/tex]
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+ Â 0.103 mol + 0.103 mol
n = 1.103 moles
At equilibrium ,mole fraction of [tex]SO_2Cl_2=\chi_1=\frac{(1-0.103) mol}{1.103mol}=0.8132[/tex]
Partial pressure of f [tex]SO_2Cl_2[/tex] equilibrium ;
[tex]p_1=P\times \chi_1[/tex] (Dalton's law of partial pressure)
[tex]p_1=1.30\times 0.8132=1.057 atm[/tex]
At equilibrium ,mole fraction of [tex]SO_2=\chi_2=\frac{(0.103) mol}{1.103mol}=0.09338[/tex]
Partial pressure of f [tex]SO_2[/tex] equilibrium ;
[tex]p_2=P\times \chi_2[/tex] (Dalton's law of partial pressure)
[tex]p_2=1.30\times 0.09338=0.1214 atm[/tex]
At equilibrium ,mole fraction of [tex]Cl_2=\chi_3=\frac{(0.103) mol}{1.103mol}=0.09338[/tex]
Partial pressure of f [tex]Cl_2[/tex] equilibrium ;
[tex]p_3=P\times \chi_3[/tex] (Dalton's law of partial pressure)
[tex]p_3=1.30\times 0.09338=0.1214 atm[/tex]
The value of the [tex]K_p[/tex] of the reaction will be;
[tex]K_p=\frac{p_2\times p_3}{p_1}=\frac{0.1214 atm\times 0.1214 atm}{1.057 atm}=0.0139[/tex]
