question_answer 21)

The reaction of cyanamide, \[N{{H}_{2}}CN(s)\], with the dioxygen was carried out in a bomb calorimeter and\[\Delta U\] was found to be -742.7 kJ/mol at 298 K. Calculate the enthalpy change for the reaction at 298 K.

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: \[\Delta H=-742.7+\frac{1}{2}\times 8.314\times {{10}^{-3}}\times 298\] \[=-741.5kJ\,mo{{l}^{-1}}\]