Which of the following reaction will have greater enthalpy and why?
Maurice has stated why equation 1 is much more exothermic than equation 2. A visualization of the processes may help you understand the details. Equation 1 involves the reaction of a carbon atom with four atoms of hydrogen to form methane in the gas phase. The bond dissociation energy (BDE) of the hydrogen molecule is +104 kcal/mol. Accordingly, the heat of formation (&Delta;Hfo) of the hydrogen atom is half this value or +52 kcal/mol. The &Delta;Hfo of the carbon atom from graphite is +171 kcal/mol. Both hydrogen and graphite are taken as 0 kcal/mol at the Standard State. The right hand side of the chart shows that the sum total of the necessary atoms is +379 kcal/mol above the Standard State. This is your starting point for equation 1. To form methane in the gas phase the reaction will be exothermic by -379 kcal/mol plus the &Delta;Hfo of methane (-18 kcal/mol) for a total of -397 kcal/mol for equation 1. On the other hand, equation 2 is nothing more than the &Delta;Hfo of methane as -18 kcal/mol below the Standard State as shown in the left hand diagram. This diagram also shows the stepwise BDE's of C-H bonds starting from methane. The same value for equation 1 is obtained, -397 kcal/mol.