Consider the reaction N2 - 3H2 - 2NH3 carried out at constant temperature and pressure- If - H and - E are the enthalpy and internal energy changes for the reaction- which of the following expression is true-

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Introduction of VSEPR

Consider the ...

Question

Consider the reaction $N_{2} + 3 H_{2} \to 2 N H_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ E$ are the enthalpy and internal energy changes for the reaction, which of the following expression is true?

Δ H

= zero

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Δ H = Δ E

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Δ H < Δ E

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Δ H > Δ E

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Solution

The correct option is A $Δ H < Δ E$
Here, temperature and pressure are constant.
So, $Δ H = Δ E + Δ n R T$ .
Number of moles in reactant = 1+3 = 4
Number of moles in product = 2
So, $Δ n = 2 - 4 = - 2 S o, Δ H = Δ E - 2 R T H e n c e, Δ H < Δ E$
So, option C is correct.

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

The correct option is A ΔH<ΔEHere, temperature and pressure are constant.So, ΔH=ΔE+ΔnRT.Number of moles in reactant = 1+3 = 4Number of moles in product = 2So, Δn=2−4=−2So,ΔH=ΔE−2RTHence,ΔH<ΔESo, option C is correct.

Consider the reaction $N_{2} + 3 H_{2} \to 2 N H_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ E$ are the enthalpy and internal energy changes for the reaction, which of the following expression is true?

The correct option is A $Δ H < Δ E$
Here, temperature and pressure are constant.
So, $Δ H = Δ E + Δ n R T$ .
Number of moles in reactant = 1+3 = 4
Number of moles in product = 2
So, $Δ n = 2 - 4 = - 2 S o, Δ H = Δ E - 2 R T H e n c e, Δ H < Δ E$
So, option C is correct.