Question

0.562 g of graphite kept in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure was burnt according to the equation: $C(gr)+O_{2}(g) \longrightarrow CO_{2}(g)$
During the reaction, temperature rises from 298 K to 298.89 K. If the heat capacity of the calorimeter and its contents is 20.7 kJ/K. What is the enthalpy change for the above reaction at 298 K and 1 atm?

Answer 1

$\Delta H\quad =\quad \Delta U\quad +\quad \Delta { n }_{ g }RT$

For the reaction

$C(gr)\quad +\quad { O }_{ 2 }(g)\quad \rightarrow \quad C{ O }_{ 2 }(g);\quad \Delta { n }_{ g }\quad =\quad 0\\\\ \Delta H\quad =\quad \Delta U\quad =\quad { q }_{ v }\\ \\ \Delta H\quad =\quad { q }_{ v }\quad =\quad { C }_{ v }(\Delta T)\\ \\=\quad 20.7\quad kJ\quad { K }^{ -1 }\quad \times \quad 0.89\quad K\\ \\=\quad 18.4\quad kJ$

As the reaction is a combustion reaction that must be an exothermic reaction.

∆H = -18.4 kJ for the combustion of 0.562 g of C (graphite)

For 1 mol of C (graphite),

$\\ \Delta H\quad =\quad -18.4\quad \times \quad \frac { 12.0 }{ 0.562 } \\ \\ \Delta H\quad =\quad -392.88\quad kJ/mol$

The enthalpy of combustion of graphite is -392.88 kJ${ mol }^{ -1 }$