Consider the reaction: N_2+3H_2\rightarrow 2NH_3  carried out at constant temperature and pressure. If \Delta H\; and \; \Delta U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

  • Option 1)

    \Delta H=0\;

  • Option 2)

    \; \Delta H=\Delta U\;

  • Option 3)

    \; \Delta H< \Delta U\;

  • Option 4)

    \; \Delta H> \Delta U\;

 

Answers (1)

As we learnt in

First law of Thermodynamics -

Energy of universe is always conserved or total energy of an isolated system is always conserved

\Delta E= q + W


 

- wherein

\Delta E= Internal Energy

q= Heat

W= work

 

 \Delta H= \Delta U + \Delta nRT\ \; for\ N_{2}+3H_{2}\rightarrow 2NH_{3}

\Delta n_{g}=2-4=-2

\therefore \Delta H=\Delta U-2RT 

or    \Delta U\Delta H+2RT

\therefore \Delta U>\Delta H


Option 1)

\Delta H=0\;

This is incorrect option

Option 2)

\; \Delta H=\Delta U\;

This is incorrect option

Option 3)

\; \Delta H< \Delta U\;

This is correct option

Option 4)

\; \Delta H> \Delta U\;

This is incorrect option

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