1
Question
Determine enthalpy of formation for H2O2(l), using the listed enthalpies of reactions:
N2H4(l)+2H2O2(l)→N2(g)+4H2O(l);
△rHo1=−818 kJ/mol
N2H4(l)+O2(g)→N2(g)+2H2O(l); △rHo2=−622 kJ/mol
H2(g)+12O2(g)→H2O(l);△fHo3=−285 kJ/mol
Determine enthalpy of formation for H2O2(l), using the listed enthalpies of reactions:
N2H4(l)+2H2O2(l)→N2(g)+4H2O(l);
△rHo1=−818 kJ/mol
N2H4(l)+O2(g)→N2(g)+2H2O(l); △rHo2=−622 kJ/mol
H2(g)+12O2(g)→H2O(l);△fHo3=−285 kJ/mol
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Solution
The correct option is C -472 kJ/mol
Reaction of formation of H2O2. can be calculated by reversing the first reaction and then subtracting from reaction (2) and adding to double of reaction 3. Since, it will give heat of formation for 2 moles we will calculate for 1 mole
N2(g)+4H2O(l)→N2H4(l)+2H2O2(l);
△rHo1=818 kJ/mol...(1)
N2H4(l)+O2(g)→N2(g)+2H2O(l); △rHo2=−622 kJ/mol ...(2)
[H2(g)+12O2(g)→H2O(l);△fHo3=−285 kJ/mol] ×2 ....(3)
For H2(g)+O2(g)→H2O2(l)
△fHo(H2O2,l)=2×△rHo3+△rHo2−△rHo1
=(2×−285)−622+818
=−374 kJ
Now, heat of formation of 1 mole = −3742 = −187 kJ/mole
Reaction of formation of H2O2. can be calculated by reversing the first reaction and then subtracting from reaction (2) and adding to double of reaction 3. Since, it will give heat of formation for 2 moles we will calculate for 1 mole
N2(g)+4H2O(l)→N2H4(l)+2H2O2(l);
△rHo1=818 kJ/mol...(1)
N2H4(l)+O2(g)→N2(g)+2H2O(l); △rHo2=−622 kJ/mol ...(2)
[H2(g)+12O2(g)→H2O(l);△fHo3=−285 kJ/mol] ×2 ....(3)
For H2(g)+O2(g)→H2O2(l)
△fHo(H2O2,l)=2×△rHo3+△rHo2−△rHo1
=(2×−285)−622+818
=−374 kJ
Now, heat of formation of 1 mole = −3742 = −187 kJ/mole
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