Concepts and reason
The change in energy ( Δ H ) \left( {\Delta H} \right) ( Δ H ) during phase change or during a reaction is known as Enthalpy. When energy is absorbed during reaction or phase change the sign for enthalpy is taken as positive and when energy is released during reaction or phase change the sign for enthalpy is taken as negative.
In the given question, you need to determine Δ H \Delta H Δ H for the combustion reaction.
Fundamentals
When an organic substance is burnt in presence of air it always produces carbon dioxide and water. This reaction is exothermic in nature as heat is evolved during the reaction.
Part A
The reaction is as follows:
2 C H 3 O H ( g ) + 3 O 2 ( g ) → 2 C O 2 ( g ) + 4 H 2 O ( g ) {\rm{2C}}{{\rm{H}}_3}{\rm{OH}}\left( g \right) + 3{{\rm{O}}_2}\left( g \right) \to 2{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right) + 4{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right) 2 C H 3 O H ( g ) + 3 O 2 ( g ) → 2 C O 2 ( g ) + 4 H 2 O ( g )
Part B
The reaction for the combustion of one mole of methanol is as follows:
C H 3 O H ( g ) + 3 2 O 2 ( g ) → C O 2 ( g ) + 2 H 2 O ( g ) {\rm{C}}{{\rm{H}}_3}{\rm{OH}}\left( g \right) + \frac{3}{2}{{\rm{O}}_2}\left( g \right) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right) + 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right) C H 3 O H ( g ) + 2 3 O 2 ( g ) → C O 2 ( g ) + 2 H 2 O ( g )
Now, the enthalpy of the reaction is calculated as follows:
Δ H r x n = B o n d e n e r g i e s o f a l l t h e r e a c t a n t s − B o n d e n e r g i e s o f a l l t h e p r o d u c t s = [ 3 ( 1 m o l ) ( C − H ) + 1 ( 1 m o l ) ( C − O ) + 1 ( 1 m o l ) ( O − H ) + 1 ( 3 2 m o l ) ( O = O ) ] − [ 2 ( 1 m o l ) ( C = O ) + 2 ( 2 m o l ) ( O − H ) ] = [ 3 ( 1 m o l ) ( 4 1 4 k J / m o l ) + 1 ( 1 m o l ) ( 3 6 0 k J / m o l ) + 1 ( 1 m o l ) ( 4 6 4 k J / m o l ) + 1 ( 3 2 m o l ) ( 4 9 8 k J / m o l ) ] − [ 2 ( 1 m o l ) ( 7 9 9 k J / m o l ) + 2 ( 2 m o l ) ( 4 6 4 k J / m o l ) ] = − 6 4 1 k J \begin{array}{c}\\\Delta {H_{{\rm{rxn}}}} = {\rm{Bond}}\;{\rm{energies}}\;{\rm{of}}\;{\rm{all}}\;{\rm{the}}\;{\rm{reactants}} - {\rm{Bond}}\;{\rm{energies}}\;{\rm{of}}\;{\rm{all}}\;{\rm{the}}\;{\rm{products}}\\\\ = \left[ {3\left( {1\;{\rm{mol}}} \right)\left( {{\rm{C}} - {\rm{H}}} \right) + 1\left( {1\;{\rm{mol}}} \right)\left( {{\rm{C}} - {\rm{O}}} \right) + 1\left( {1\;{\rm{mol}}} \right)\left( {{\rm{O}} - {\rm{H}}} \right) + 1\left( {\frac{3}{2}\;{\rm{mol}}} \right)\left( {{\rm{O}} = {\rm{O}}} \right)} \right] - \\\\\left[ {2\left( {1\;{\rm{mol}}} \right)\left( {{\rm{C}} = {\rm{O}}} \right) + 2\left( {2\;{\rm{mol}}} \right)\left( {{\rm{O}} - {\rm{H}}} \right)} \right]\\\\ = \left[ {3\left( {1\;{\rm{mol}}} \right)\left( {414\;{\rm{kJ/mol}}} \right) + 1\left( {1\;{\rm{mol}}} \right)\left( {360\;{\rm{kJ/mol}}} \right) + 1\left( {1\;{\rm{mol}}} \right)\left( {{\rm{464}}\;{\rm{kJ/mol}}} \right) + 1\left( {\frac{3}{2}\;{\rm{mol}}} \right)\left( {498\;{\rm{kJ/mol}}} \right)} \right] - \\\\\left[ {2\left( {1\;{\rm{mol}}} \right)\left( {799\;{\rm{kJ/mol}}} \right) + 2\left( {2\;{\rm{mol}}} \right)\left( {464\;{\rm{kJ/mol}}} \right)} \right]\\\\ = - 641\;{\rm{kJ}}\\\end{array} Δ H r x n = B o n d e n e r g i e s o f a l l t h e r e a c t a n t s − B o n d e n e r g i e s o f a l l t h e p r o d u c t s = [ 3 ( 1 m o l ) ( C − H ) + 1 ( 1 m o l ) ( C − O ) + 1 ( 1 m o l ) ( O − H ) + 1 ( 2 3 m o l ) ( O = O ) ] − [ 2 ( 1 m o l ) ( C = O ) + 2 ( 2 m o l ) ( O − H ) ] = [ 3 ( 1 m o l ) ( 4 1 4 k J / m o l ) + 1 ( 1 m o l ) ( 3 6 0 k J / m o l ) + 1 ( 1 m o l ) ( 4 6 4 k J / m o l ) + 1 ( 2 3 m o l ) ( 4 9 8 k J / m o l ) ] − [ 2 ( 1 m o l ) ( 7 9 9 k J / m o l ) + 2 ( 2 m o l ) ( 4 6 4 k J / m o l ) ] = − 6 4 1 k J
Part C
The reaction is as follows:
N 2 ( g ) + 3 H 2 ( g ) → 2 N H 3 ( g ) {{\rm{N}}_2}\left( g \right) + 3{{\rm{H}}_2}\left( g \right) \to 2{\rm{N}}{{\rm{H}}_3}\left( g \right) N 2 ( g ) + 3 H 2 ( g ) → 2 N H 3 ( g )
Now, the enthalpy of the reaction is calculated as follows:
Δ H r x n = B o n d e n e r g i e s o f a l l t h e r e a c t a n t s − B o n d e n e r g i e s o f a l l t h e p r o d u c t s = [ 1 ( 1 m o l ) ( N ≡ N ) + 1 ( 3 m o l ) ( H − H ) ] − [ 3 ( 2 m o l ) ( N − H ) ] = [ 1 ( 1 m o l ) ( 9 4 6 k J / m o l ) + 1 ( 3 m o l ) ( 4 3 6 k J / m o l ) ] − [ 3 ( 2 m o l ) ( 3 8 9 k J / m o l ) ] = − 8 0 k J \begin{array}{c}\\\Delta {H_{{\rm{rxn}}}} = {\rm{Bond}}\;{\rm{energies}}\;{\rm{of}}\;{\rm{all}}\;{\rm{the}}\;{\rm{reactants}} - {\rm{Bond}}\;{\rm{energies}}\;{\rm{of}}\;{\rm{all}}\;{\rm{the}}\;{\rm{products}}\\\\ = \left[ {1\left( {1\;{\rm{mol}}} \right)\left( {{\rm{N}} \equiv {\rm{N}}} \right) + 1\left( {3\;{\rm{mol}}} \right)\left( {{\rm{H}} - {\rm{H}}} \right)} \right] - \left[ {3\left( {2\;{\rm{mol}}} \right)\left( {{\rm{N}} - {\rm{H}}} \right)} \right]\\\\ = \left[ {1\left( {1\;{\rm{mol}}} \right)\left( {946\;{\rm{kJ/mol}}} \right) + 1\left( {3\;{\rm{mol}}} \right)\left( {436\;{\rm{kJ/mol}}} \right)} \right] - \left[ {3\left( {2\;{\rm{mol}}} \right)\left( {389\;{\rm{kJ/mol}}} \right)} \right]\\\\ = - 80\;{\rm{kJ}}\\\end{array} Δ H r x n = B o n d e n e r g i e s o f a l l t h e r e a c t a n t s − B o n d e n e r g i e s o f a l l t h e p r o d u c t s = [ 1 ( 1 m o l ) ( N ≡ N ) + 1 ( 3 m o l ) ( H − H ) ] − [ 3 ( 2 m o l ) ( N − H ) ] = [ 1 ( 1 m o l ) ( 9 4 6 k J / m o l ) + 1 ( 3 m o l ) ( 4 3 6 k J / m o l ) ] − [ 3 ( 2 m o l ) ( 3 8 9 k J / m o l ) ] = − 8 0 k J
Ans: Part A
The balanced combustion reaction of methanol is as follows:
2 C H 3 O H ( g ) + 3 O 2 ( g ) → 2 C O 2 ( g ) + 4 H 2 O ( g ) {\rm{2C}}{{\rm{H}}_3}{\rm{OH}}\left( g \right) + 3{{\rm{O}}_2}\left( g \right) \to 2{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right) + 4{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right) 2 C H 3 O H ( g ) + 3 O 2 ( g ) → 2 C O 2 ( g ) + 4 H 2 O ( g )
Part B
The enthalpy of the reaction is − 6 4 1 k J - 641\;{\rm{kJ}} − 6 4 1 k J for one mole of methanol.
Part C
The enthalpy of the reaction is − 8 0 k J - 80\;{\rm{kJ}} − 8 0 k J .