Question

The reaction of cyanamide, $NH_{2}CN$ (s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ $mol^{-1}$ at 298 K. Calculate enthalpy change for the reaction at 298 K. $NH_{2}CN(s)+\frac{3}{2}O_{2}(g) \longrightarrow N_{2}(g)+CO_{2}(g)+H_{2}O(l)$

Answer 1

The given reaction is as follows.

            ${ NH }_{ 2 }CN(g)\quad +\quad \frac { 3 }{ 2 } { O }_{ 2 }(g)\quad \rightarrow\quad { N }_{ 2 }(g)\quad +\quad { CO }_{ 2 }(g)\quad +\quad { H }_{ 2 }O$

From the given,

           $\Delta U\quad =\quad -742.7\quad KJ.{ mol }^{ -1 }$

From the reaction

Number of moles of product = 2

Number of moles of reactants = $\frac { 3 }{ 2 }$

          $\Delta ng\quad ={ \quad n }_{ p }\quad -\quad { n }_{ r }$

          $=\quad 2\quad -\quad \frac { 3 }{ 2 }$

          $\Delta ng\quad =\quad +\frac { 1 }{ 2 } \quad mol$

          $R\quad =\quad 8.314\quad \times \quad { 10 }^{ -3 }\quad KJ.{ mol }^{ -1 }$

We know,

          $\Delta \Eta \quad =\quad -742.7\quad KJ.{ mol }^{ -1 }\quad +\quad \left( \frac { 1 }{ 2 } mol \right) \quad \times \quad \left( 8.34\times { 10 }^{ -3 }KJ.{ K }^{ -1 }.{ mol }^{ -1 } \right) \quad \times \quad \left( 298\quad K \right)$

          $=\quad \left( -742.7\quad +\quad 1.2 \right) KJ.mo{ l }^{ -1 }$

          $\Delta \Eta \quad =\quad 741.5\quad KJ.{ mol }^{ -1 }$

Answer 2

Heya mate☺☺
∆=741.5 kJ/mol