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We are asked to calculate ΔH°f of N2H4(aq).
2 NH3(aq) + NaOCl(aq) ⟶ N2H4(aq) + NaCl(aq) + H2O(l) ΔH°rxn = −151 kJ
Gather ΔH°f values from the standard table:
ΔH°f of NaOCl(aq) = −346 kJ/mol
ΔH°f of NH3(aq) = −80.83 kJ/mol
ΔH°f of NaCl(aq) = −407 kJ/mol
ΔH°f of H2O(aq) = −285.8 kJ/mol
We can use the following equation to solve for ΔH˚rxn:
Note that we need to multiply each ΔH˚f by the stoichiometric coefficient since ΔH˚f is in kJ/mol.
Also, note that ΔH˚f for elements in their standard state is 0.
Electric generating plants transport large amounts of hot water through metal pipes, and oxygen dissolved in the water can cause a major corrosion problem. Hydrazine (N2H4) added to the water prevents this problem by reacting with the oxygen:
N2H4(aq) + O2(g) ⟶ N2(g) + 2H2O(l)
About 4 x 107 kg of hydrazine is produced every year by reacting ammonia with sodium hypochlorite in the Raschig process:
2NH3(aq) + NaOCl(aq) ⟶ N2H4(aq) + NaCl(aq) + H2O(l) ΔH°rxn = −151 kJ
If ΔH°f of NaOCl(aq) = −346 kJ/mol, find ΔH°f of N2H4(aq).
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Based on our data, we think this problem is relevant for Professor Arman's class at UTSA.
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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.