Answer:
(a) Kc = 264
(b) Kp = 6.15
Explanation:
(a) To calculate Kc we need the molar concentrations of each substance.
[tex]M=\frac{molesofsolute}{litresofsolution} =\frac{massofsolute}{molarmassofsolute \times litresofsolution}[/tex]
[tex][PCl_{3}]=\frac{0.220g}{137.5g/mol \times 25.0L } =6.40 \times 10^{-5} M[/tex]
[tex][Cl_{2}]=\frac{2.12g}{71.0g/mol \times 25.0L } =1.19 \times 10^{-3}M[/tex]
[tex][PCl_{5}]=\frac{0.105g}{208.5g/mol \times 25.0L } =2.01 \times 10^{-5} M[/tex]
Then, we replace these concentrations in the Kc expression.
[tex]Kc=\frac{[PCl_{5}]}{[PCl_{3}]\times [Cl_{2}] } =\frac{2.01 \times 10^{-5} }{6.40 \times 10^{-5} \times 1.19 \times 10^{-3} } =264[/tex]
(b) To find out Kp we can use the following expression:
[tex]Kp = Kc.(R.T)^{\Delta n(g) }[/tex]
where,
R is the ideal gas constant (0.08206 atm .L /mol . K)
T is the absolute temperature (250°C + 273.15 = 523.15 K)
Δn is gaseous moles of products - gaseous moles of reactants (1 - 2 = -1)
If we replace with the values we have:
[tex]Kp = Kc.(R.T)^{\Delta n(g) }=264 \times (0.08206 \times 523.15)^{-1} =6.15[/tex]
