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Question
Given that: C(s)+O2(g)→CO2(g); ΔH=−394 kJ and 2H2(g)+O2(g)→2H2O(l),ΔH=−568 kJ and C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l); ΔH=−1058 kJ/mole. Using the data, the heat of formation of ethanol is:
Given that: C(s)+O2(g)→CO2(g); ΔH=−394 kJ and 2H2(g)+O2(g)→2H2O(l),ΔH=−568 kJ and C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l); ΔH=−1058 kJ/mole. Using the data, the heat of formation of ethanol is:
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Solution
The correct option is A −582 kJ
Solution:-C(s)+O2(g)⟶CO2(g)ΔH=−394KJ/mol2×[C(s)+O2(g)⟶CO2(g)ΔH=−394KJ/mol]2C(s)+2O2(g)⟶2CO2(g)ΔH=−788KJ/mol.....(1)2H2(g)+O2(g)⟶2H2O(l)ΔH=−568KJ/mol32×[2H2(g)+O2(g)⟶2H2O(l)ΔH=−852KJ/mol]
3H2(g)+32O2(g)⟶3H2O(l)ΔH=−568KJ/mol.....(2)
C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)ΔH=−1058KJ/mol2CO2(g)+3H2O(l)⟶C2H5OH(l)+3O2(g)ΔH=1058KJ/mol.....(3)Adding eqn(1),(2)&(3), we have2C(s)+2O2(g)+3H2(g)+32O2(g)+2CO2(g)+3H2O(l)⟶2CO2(g)+3H2O(l)+C2H5OH(l)+3O2(g)2C(s)+3H2(g)+12O2(g)⟶C2H5OH(l)ΔH=−582KJ/molHence the heat of formation of ethanol is −582KJ/mol.
Solution:-
C(s)+O2(g)⟶CO2(g)ΔH=−394KJ/mol
2×[C(s)+O2(g)⟶CO2(g)ΔH=−394KJ/mol]
2C(s)+2O2(g)⟶2CO2(g)ΔH=−788KJ/mol.....(1)
2H2(g)+O2(g)⟶2H2O(l)ΔH=−568KJ/mol
32×[2H2(g)+O2(g)⟶2H2O(l)ΔH=−852KJ/mol]
3H2(g)+32O2(g)⟶3H2O(l)ΔH=−568KJ/mol.....(2)
C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)ΔH=−1058KJ/mol
2CO2(g)+3H2O(l)⟶C2H5OH(l)+3O2(g)ΔH=1058KJ/mol.....(3)
Adding eqn(1),(2)&(3), we have
2C(s)+2O2(g)+3H2(g)+32O2(g)+2CO2(g)+3H2O(l)⟶2CO2(g)+3H2O(l)+C2H5OH(l)+3O2(g)
2C(s)+3H2(g)+12O2(g)⟶C2H5OH(l)ΔH=−582KJ/mol
Hence the heat of formation of ethanol is −582KJ/mol.
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