Q

# How to solve this problem- - Chemical Thermodynamics - JEE Main

Consider the reaction: $\dpi{100} N_2+3H_2\rightarrow 2NH_3$  carried out at constant temperature and pressure. If $\dpi{100} \Delta H\; and \; \Delta U$ 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)

$\; \Delta H> \Delta U\;$

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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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