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For the reaction at 25^(@)C NH(3(g))rarr...

For the reaction at `25^(@)C` `NH_(3(g))rarr(1)/(2)N_(2(g))+(3)/(2)H_(2(g)),DeltaH^(@)=11.04K` calculate `DeltaU^(@)` of the reaction at the given temperature

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To calculate the change in internal energy (\( \Delta U \)) for the reaction: \[ \text{NH}_3(g) \rightarrow \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \] given that the enthalpy change (\( \Delta H \)) at \( 25^\circ C \) is \( 11.04 \, \text{kcal} \), we can use the relation between enthalpy change and internal energy change: \[ \Delta H = \Delta U + \Delta N_g R T \] ...
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