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 avoids the problem by reacting with the oxygen:
N₂H₄(aq) + O₂(g) → N₂(g) +2H₂O(l)About 4X10⁷ 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 = -151kj
(b) What is the heat released when aqueous N₂H₄ is added to 5.00X10³ L of water that is 2.50X10⁻⁴ M O₂?

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Respuesta :

shaikhmohd0119 shaikhmohd0119 · Answer 1 · 2022-09-04T16:46:04+02:00

Heat Released when aqueous N₂H₄ is added to 5.00X10³ L of water that is 2.50X10⁻⁴ M O₂ is 1.02438 X10³ kJ

Now,

For the given reaction,

N₂H₄(aq) + O₂(g) → N₂(g) +2H₂O(l)

ΔH(reaction) = ∑H(product) - ∑H(reactants)

where,

∑H(product) = Summation of heat of formation of products

∑H(reactants)= Summation of heat of formation of reactants

ΔrH° = [ ΔfH°(N₂) + 2×ΔfH°(H₂0)] - [ΔfH°(N₂H₂) + ΔfH°(O₂)]

ΔrH° = [0+2× -285.84]-[35.36+0]

= -607.04 kJ/mol

Now,

According to the question,

In 1L of water moles of oxygen present = 2.5×10⁻⁴

In 6.75 ×10³ L of water, moles of oxygen present = 2.5×10⁻⁴×6.75 ×10³ moles of oxygen

Now,

According to the reaction,

For 1 mole of O₂, heat released = -607.04kJ

For 1.6875 moles of O₂ heat released =-607.04 × 1.6875 KJ= 1024.38

= 1.02438×10³ kJ

From the above conclusion we can say that ,Heat Released when aqueous N₂H₄ is added to 5.00X10³ L of water that is 2.50X10⁻⁴ M O₂ is 1.02438 ×10³ kJ

Learn more about Raschig process here: https://brainly.com/question/17924325

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