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enthalpy question

How would you do the question below?

The Born-Haber process, Reaction Q2.b has an enthalpy change of
– 92.4 kJ mol−1 at standard conditions of 298 K and 105 Pa.

N2 + 3H2 → 2NH3

Reaction Q2.b

Assuming that the gases behave as ideal gases throughout and the pressure remains constant, calculate the enthalpy change of the reaction at the typical operating conditions of 670 K.
Original post by gamma^infinity
How would you do the question below?

The Born-Haber process, Reaction Q2.b has an enthalpy change of
– 92.4 kJ mol−1 at standard conditions of 298 K and 105 Pa.

N2 + 3H2 → 2NH3

Reaction Q2.b

Assuming that the gases behave as ideal gases throughout and the pressure remains constant, calculate the enthalpy change of the reaction at the typical operating conditions of 670 K.


You can use the enthalpy change to calculate the internal energy change of the system using:

ΔH = ΔU + PΔV

The change in volume at the conditions given = 4 mol --> 2 mol, which at standard conditions = -(2 x 22.7 dm3) = -45.4 dm3 = -0.0454 m3

ΔH = ΔU + PΔV

-92400 J = ΔU - 0.0454 x 105

ΔU = -87.86 kJ mol

Now do the reverse for the new conditions:

volume of 2 mol gas at 105 Pa and 670 K (constant pressure)

V1/T1 = V2/T2

V2 = V1T2/T1 = (0.0454 x 670)/273 = 0.1114 m3

ΔH = -87860 - 0.1114 x 105

ΔH = 99.0 kJ
(edited 5 years ago)

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