1 gram of graphite is burnt in a bomb calorimeter in excess of oxygenat 298 K and 1 atm pressure according to the equation : C (graphite) + O2 (g) → CO2 (g).During the reaction temperature rises from 298 to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K then what is the enthalpy change for the above reaction at 298 K and 1 atm?
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Given:
\(C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}\)
\(\Delta n_g=1-1=0\)
weight of C =1
T1=298
T2=299
Cv=20.7 KJ/k
\(\Delta h=?\)
\(mol=\frac{1}{12}\)
We know
\(\Delta H=\Delta E+\Delta n_g RT\)
Where \(\Delta n_g=0\)
So, \(\Delta H=\Delta E\) ........(1)
Heat evolved at constant volume
\(q_v= - C_v \Delta T=-20.7\times(299-298)\\=-20.7\times1=-20.7KJ\)
\(\frac{q_v}{n}=\Delta E\)(for 1mol)
\(\Delta E=\frac{-20.7}{\frac{1}{12}}\\=-20.7\times12\)
\(\Delta E=-20.7\times12\,KJ/mol\)
From equation (1)
\(\Delta H=\Delta E=-20.7\times12\,KJ/mol\\=-248.4\,KJ/mol\)
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