Consider the reaction: N_2(g) + 3H_2(g) → 2NH_3(g) carried out at constant temperature and pressure. If H and U are the enthalpy and internal energy c - Chemistry MCQ AIEEE 2026 | ExamTest.live

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Consider the reaction: N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g) carried out at constant temperature and pressure. If Δ H and Δ U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

Solution & Step-by-step Explanation

The relationship between Δ H and Δ U is: Δ H = Δ U + Δ n_{g} RT For the reaction N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g) : Δ n_{g} = (moles of gaseous products) - (moles of gaseous reactants) Δ n_{g} = 2 - (1 + 3) = - 2 Substituting this: Δ H = Δ U - 2 RT Since R and T are positive, Δ H is obtained by subtracting a positive value from Δ U .Thus, Δ H < Δ U .

Try it yourself before checking the explanation above.

Consider the reaction: N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g) carried out at constant temperature and pressure. If Δ H and Δ U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

Δ H = 0

Δ H = Δ U

Δ H < Δ U

Δ H > Δ U

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