1. Tardigrade
  2. Question
  3. Chemistry
  4. The enthalpy change (Δ H ) for the reaction, N 2( g )+3 H 2( g ) longrightarrow 2 NH 3( g ) is -92.38 kJ at 298 K. The internal energy change Δ U at 298 K is:

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Q. The enthalpy change $(\Delta H )$ for the reaction, $N _{2}( g )+3 H _{2}( g ) \longrightarrow 2 NH _{3}( g )$ is $-92.38 \,kJ$ at $298 \,K$. The internal energy change $\Delta U$ at $298 K$ is:

AIIMSAIIMS 2006Thermodynamics

Solution:

$\Delta H =\Delta E + P \Delta V$
$ \Delta H =\Delta E +\Delta nRT $
$ \Delta E =\Delta H -\Delta nRT $
$=-92.38-(-2)(8.314) \times 10^{-3} \times(298) $
$ \Delta E =-92.38+4.895 $
$=-87.48 \,kJ$