Question 1
For the reaction SbCl
5(g)

SbCl
3(g) + Cl
2(g),
ΔG°
f (SbCl
5) = –334.34 kJ/mol
ΔG°
f (SbCl
3) = –301.25 kJ/mol
ΔH°
f (SbCl
5) = –394.34 kJ/mol
ΔH°
f (SbCl
3) = –313.80 kJ/mol
Calculate the value of the equilibrium constant (K
p) for this reaction at 298 K.
Question 2
For the reaction SbCl
5(g)

SbCl
3(g) + Cl
2(g),
ΔG°
f (SbCl
5) = –334.34 kJ/mol
ΔG°
f (SbCl
3) = –301.25 kJ/mol
ΔH°
f (SbCl
5) = –394.34 kJ/mol
ΔH°
f (SbCl
3) = –313.80 kJ/mol
Calculate the value of the equilibrium constant (K
p) at 800 K and 1 atm pressure.