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6.8  The reaction of cyanamide,NH_2 CN (s), with dioxygen was carried out in a bomb calorimeter, and \Delta U was found to be -742.7 KJ mol ^{-1} at 298 K. Calculate enthalpy change for the reaction at 298 K.

         NH_2 CN ( g ) + \frac{3}{2} O_2 (g) \rightarrow N_2 + CO_2 ( g ) H_2 O ( l)

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Given information,
\Delta U = -742.7kJmol^{-1}
T = 298 K
R = 8.314\times 10^{-3}

\Delta n_g =n_g(products)- n_g(reactants)
           = (2 - 1.5)
           = 0.5 moles

The enthalpy change for the reaction is expressed as;

 \Delta H = \Delta U+n_gRT
where, \Delta U = change in internal energy  and
 \Delta n_g = change in no. of moles

By putting the values we get,
\Delta H = (-742.7)+0.5(298)(8.314\times 10^{-3})
           =-741.5 kJmol^{-1}

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manish

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