Answer:
Information shadow:
The given reaction
is:
\[N{{H}_{2}}CN(s)+\frac{3}{2}{{O}_{2}}(g)\to
{{N}_{2}}(g)+C{{O}_{2}}(g)+{{H}_{2}}O(l)\]\[\Delta U=-742.7kJ/mol\,at\,298K\]
\[\Delta
{{n}_{g}}\] = Number of gaseous moles of products
- Number of
gaseous moles of reactants
\[=2-\frac{3}{2}=\frac{1}{2}\]
Problem
solving strategy:
Enthalpy
change \[\Delta H\] can be calculated as,
\[\Delta H=\Delta
U+\Delta {{n}_{g}}RT\]
Working
it out:
\[\Delta
H=-742.7+\frac{1}{2}\times 8.314\times {{10}^{-3}}\times 298\]
\[=-741.5kJ\,mo{{l}^{-1}}\]
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