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Given `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g),Delta_(r)H^(Ө)= -92.4 kJ mol^(-1)` What is the standard enthalpy of formation of `NH_(3)` gas? |
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Answer» Correct Answer - `-46.2 kJ mol^(-1)` Standard enthalpy of formation of a compound is the change in ethalpy that takes place during the formation of 1 mole of a substance in its standrad from its constituent elements in their standrad state. Re-writing the given equation for 1 mole of `MH_(3(g))` `(1)/(2)N_(2(g))+(3)/(2)H_(2(g))rarrNH_(3(g))` `:.` Standard enthalpy of formation of `NH_(3(g))` `=1//2Delta_(r)H^(theta)` `1//2 (-92.4 kJ "mol"^(-1))` `=-46.2 kJ "mol"^(-1)` |
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