Question

Consider the reaction : N₂ + 3H₂ → 2NH₃ carried out at
constant temperature and pressure. If ∆H and ∆U are the
enthalpy and internal energy changes for the reaction, which
of the following expressions is true ?
(a) ∆H > ∆U (b) ∆H < ∆U
(c) ∆H = ∆U (d) ∆H = 0

Answer 1

The correct expression is  ∆H < ∆U

Option (B) is correct.

Explanation:

N₂ + 3H₂ → 2NH₃

∆H = Change in enthalpy

∆U = Change in internal energy

∆H  = ∆U + ∆ng RT

Now ∆ng can be calculated as:

∆ng = Product moles - reactant moles

∆ng = 2 - 4 = - 2

Now

∆H  = -2 RT

So ∆H  < ∆U

Thus the correct expression is  ∆H  < ∆U

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