a>
đơn vị [tex]K_{p}[/tex] là atm vì khi cân bằng
[tex]K_{p}[/tex] = ([tex]P_{SO_{2}}*P_{Cl_{2}}[/tex]) / ([tex]P_{SO_{2}Cl_{2}}[/tex])=50(atm*atm)/(atm)=50atm
b>đặt [tex]n_{SO_{2}Cl_{2}}[/tex] = 1 mol
[tex]SO_{2}Cl_{2}[/tex] (k) [tex]\rightleftharpoons[/tex] [tex]SO_{2}_{k} + Cl_{2}_{k}[/tex] (1) [tex]K_{p}[/tex] =50
ban đầu 1 mol 0 0
phân li x mol x mol x mol
[ ] (1-x) mol x mol x mol
[tex]\sum n[/tex] = (1+x ) mol [tex]\Leftrightarrow[/tex] P= 2 atm
[tex]n_{Cl_{2}} = n_{SO_{2}}[/tex] [tex]\Leftrightarrow[/tex] [tex]P_{Cl_{2}} = P_{SO_{2}}[/tex] =[tex]\frac{2x}{1+x}[/tex] atm
[tex]n_{SO_{2}Cl_{2}}[/tex] =(1-x) [tex]\Leftrightarrow[/tex] [tex]P_{SO_{2}Cl_{2}}[/tex] = [tex]\frac{(1-x)^{2}}{1+x}[/tex] atm
[tex]k_{p}[/tex] = [tex]\frac{P_{Cl_{2}}*P{_{SO_{2}}}}{P_{SO_{2}Cl_{2}}}[/tex] = [tex]\frac{(\frac{2x}{1+x})^{2}}{\frac{(1-x)^{2}}{1+x}}[/tex]=50
[tex]\Rightarrow[/tex] x = [tex]\sqrt{\frac{50}{52}}[/tex] = 0,9806
số mol [tex]SO_{2}Cl_{2}[/tex] còn lại = 1 - 0,9806= 0,0194
[tex]\Rightarrow[/tex] %[tex]V_{SO_{2}Cl_{2}}[/tex] = [tex]\frac{0,0194*100}{1+0,9806}[/tex] =0,98%
c> [tex]n_{Cl_{2}}=n_{SO_{2}}=x mol =n_{SO_{2}Cl_{2}}^{phân li}[/tex]
=150*0,9806=147,09