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Consider the reaction: N2 + 3H2 → 2NH3, carried out at constant temperature and pressure. If ΔH and ΔU are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

1. ΔH > ΔU
2. ΔH < ΔU
3. ΔH = ΔU
4. ΔH = 0

Correct answer: ΔH < ΔU

Solution

For N2 + 3H2 -> 2NH3, delta n_gas = 2 - 4 = -2. Since delta H = delta U + (delta n_gas)RT = delta U - 2RT, we get delta H < delta U.

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