1. Tardigrade
  2. Question
  3. Chemistry
  4. 1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation, C textgraphite)+O2(g) → CO2(g) During the reaction, temperature rises from 298 K to 299 K If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm ?

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Q. $1 \,g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298\, K$ and $1$ atmospheric pressure according to the equation,
$C_{\text{graphite)}}+O_{2(g)} \to CO_{2(g)}$
During the reaction, temperature rises from $298 \,K$ to $299\, K$ If the heat capacity of the bomb calorimeter is $20.7 \,kJ/K$, what is the enthalpy change for the above reaction at $298 \,K$ and $1\, atm$ ?

Thermodynamics

Solution:

Heat evolved in the reaction $=C_{v}\,\Delta T$
$=-20.7 \times 1 = -20.7\,kJ$
(Negative value indicates the exothermic nature of the reaction)
Heat evolved in the combustion of 1 mole of graphite, i.e.$ 12 \,g$ graphite will be
$\Delta U =-20.7 \times 12\,kJ\,mol^{-1}$
$=-2.48 \times 10^{2}\,kJ\,mol^{-1}$
$\Delta H = \Delta U+\Delta_{g}RT$
$\Delta \,n_{g}=0$ for the given reaction
$\therefore \Delta H= \Delta U =-2.48 \times 10^{2} KJ\,mol^{-1}$