Question

To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at 500°C. At equilibrium he found the concentration of HI which is equal to 0.05 M. Calculate K_c and K_p.

Answer 1

V = 3 L

`["HI"]_"initial" = (0.3 "mol")/(3"L")` 0.1 M

[HI]_eq = 0.05 M

\[\ce{2 HI(g) <=> H2(g) + l_2(g)}\]

	HI (g)	H2 (g)	I2 (g)
Initial concentration	0.1	-	-
Reacted	0.05	-	-
Equilibrium concentration	0.05	0.025	0.025

`"K"_"C" =(["H"_2]["I"_2])/(["HI"]^2)`

`= (0.025 xx 0.025)/(0.05 xx 0.05)` = 0.25

K_p = K_c(RT)^(∆n_g)

∆n_g = 2 – 2 = 0

K_p = 0.25(RT)⁰

K_p = 0.25.