Question

The equilibrium constant K_p for the reaction \[\ce{N2 (g) + 3H2 (g) <=> 2NH3 (g)}\] is 8.19 × 10² at 298 K and 4.6 × 10^-1 at 498 K. Calculate ∆H° for the reaction.

Answer 1

K_p1 = 8.19 × 10²;

T₁ = 298 K

K_p1 = 8.19 × 10²;

T₁ = 298 K

K_p2 = 4.16 × 10^-1;

T₂ = 498 K

`log  (("K"_("P"_2))/("K"_("P"_1))) = (Delta "H"^0)/(2.303  "R") [("T"_2 - "T"_1)/("T"_2"T"_1)]`

`log ((4.6 xx 106-1)/(8.19 xx 10^2)) = (Delta "H"^0)/(2.303 xx 8.314)`

`= ((498 - 298)/(498 xx 298))`

`((- 3.2505 xx 2.303 xx 8.314 xx 498 xx 298)/200) = Delta "H"^0`

ΔH⁰ = - 46181 J mol^-1

ΔH⁰ = - 46.18 KJ mol^-1