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# Consider the reaction ${{N}_{2}}(g)+3{{H}_{2}}(g)->2N{{H}_{3}}(g)$carried out at constant pressure and temperature. If $\mathbf{\Delta }H$and $\mathbf{\Delta }U$are enthalpy change and internal energy change respectively, then which of the following expressions is true?A.$\mathbf{\Delta }H=0$B.$\mathbf{\Delta }H=\mathbf{\Delta }U$C.$\mathbf{\Delta }H<\mathbf{\Delta }U$D.$\mathbf{\Delta }H>\mathbf{\Delta }U$

Last updated date: 19th Jun 2024
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Hint: Try to find out the difference in the number of moles of product and reactants. Then try to put it in an equation which relates $\Delta H,\Delta U$ and number of moles, n. An analysis can be drawn afterwards, which will lead to the answer.

${{N}_{2}}+3{{H}_{2}}->2N{{H}_{3}}$ (all are in gaseous form)
Let $\Delta n$be the difference of moles between products and reactants.
So we obtain $\Delta n$= 2-(3+1)= -2
$\mathbf{\Delta }H-\mathbf{\Delta }U+\mathbf{\Delta }nRT$, we have ( R=universal gas constant , T= temperature)
$\mathbf{\Delta }H=\mathbf{\Delta }U-2RT$, as $\Delta n$=2
Now let us observe the equation. The values of R are constant and T is also constant as it is given in the question. That means, the term ‘2RT’ is a positive term. When a positive term is getting subtracted from a bigger positive number i.e $\Delta U$ we obtain$\Delta H$, which is also positive. So, $\Delta H<\Delta U$is the required condition which gives option C as the answer.