The response of cyanamide, NH2CN(s), with dioxygen was done in a bomb calorimeter, and ∆U was observed to be – 742.7 kJ mol–1 at 298 K. Ascertain enthalpy change for the response at 298 K.

    \[NH2CN\left( g \right)\text{ }+\text{ }3/2\text{ }O2\left( g \right)\text{ }\to \text{ }N2\left( g \right)\text{ }+\text{ }CO2\left( g \right)\text{ }+\text{ }H2O\left( l \right)\]

The response of cyanamide, NH2CN(s), with dioxygen was done in a bomb calorimeter, and ∆U was observed to be – 742.7 kJ mol–1 at 298 K. Ascertain enthalpy change for the response at 298 K.

    \[NH2CN\left( g \right)\text{ }+\text{ }3/2\text{ }O2\left( g \right)\text{ }\to \text{ }N2\left( g \right)\text{ }+\text{ }CO2\left( g \right)\text{ }+\text{ }H2O\left( l \right)\]

Solution:

Enthalpy change for a response (∆H) is given by the articulation,

 

    \[H\text{ }=\text{ }U\text{ }+\text{ }ngRT\]

Where,

 

    \[\begin{array}{*{35}{l}} U\text{ }=\text{ }change\text{ }in\text{ }inside\text{ }energy  \\ ~  \\ ng\text{ }=\text{ }change\text{ }in\text{ }number\text{ }of\text{ }moles\text{ }For\text{ }the\text{ }given\text{ }response,  \\ ~  \\ ng\text{ }=\text{ }\sum ng\text{ }\left( items \right)\text{ }\text{ }\sum ng\text{ }\left( reactants \right)\text{ }=\text{ }\left( 2\text{ }\text{ }2.5 \right)\text{ }moles  \\ ~  \\ ng\text{ }=\text{ }\text{ }0.5\text{ }moles\text{ }And,\text{ }U\text{ }=\text{ }\text{ }742.7\text{ }kJ\text{ }mol1  \\ ~  \\ T\text{ }=\text{ }298\text{ }K  \\ ~  \\ R\text{ }=\text{ }8.314\text{ }\times \text{ }103\text{ }kJ\text{ }mol1\text{ }K\text{ }\text{ }1  \\ \end{array}\]

Subbing the qualities in the statement of ∆H:

    \[\begin{array}{*{35}{l}} H\text{ }=\text{ }\left( \text{ }742.7\text{ }kJ\text{ }mol1\text{ } \right)\text{ }+\text{ }\left( \text{ }0.5\text{ }mol \right)\text{ }\left( 298\text{ }K \right)\text{ }\left( 8.314\text{ }\times \text{ }103\text{ }kJ\text{ }mol1\text{ }K\text{ }\text{ }1\text{ } \right)  \\ ~  \\ =\text{ }\text{ }742.7\text{ }\text{ }1.2  \\ ~  \\ H\text{ }=\text{ }\text{ }743.9\text{ }kJ\text{ }mol1  \\ \end{array}\]