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Question
The equilibrium constant Kp 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.
Solution
Kp1 = 8.19 × 102;
T1 = 298 K
Kp1 = 8.19 × 102;
T1 = 298 K
Kp2 = 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
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