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Q. $2\, g$ of carbon is burnt in a bomb calorimeter in excess of oxygen at $298\, K$ and $1$ atmospheric pressure. During the reaction, temperature rises from $298\, K$ to $300\, K$. 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 $1\, atm$ ? $C$ (graphite) $+ O _2( g ) \rightarrow CO _2( g )$

Thermodynamics

Solution:

As we know that, $q =- C _{ V } \times \Delta T$
. $=-20.7 \times(300-298)=-41.4 \,kJ$
For combustion of $1 mol$ of graphite
$=\frac{12.0 \,g / mol \times(-41.4)}{1} $
$=-4.96 \times 10^2\, kJ / mol , \text { since } \Delta n _{ g }=0$
$\Delta H =\Delta E =-4.96 \times 10^2\, kJ / mol$