Electric generating plants transport large amounts of hot

Chapter 6, Problem 6.109

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QUESTION:

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: About 4 10 7 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) H rxn ? ?151 kJ (a) If H f of NaOCl(aq) ? ? 346 kJ/mol, find H f of N2H4(aq). (b) What is the heat released when aqueous N2H4 is added to 5.00 10 3 L of plant water that is 2.50 10 4 M O2?

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QUESTION:

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: About 4 10 7 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) H rxn ? ?151 kJ (a) If H f of NaOCl(aq) ? ? 346 kJ/mol, find H f of N2H4(aq). (b) What is the heat released when aqueous N2H4 is added to 5.00 10 3 L of plant water that is 2.50 10 4 M O2?

ANSWER:

Problem 6.109

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  added to the water avoids the problem by reacting with the oxygen:

N2H4(aq) =O2(g)\rightarrow N2(g)+2H2O(l)

About 4 x 10 7 kg of hydrazine is produced every year by reacting ammonia with sodium hypochlorite in the Raschig process:

 

                                                        \Delta H^{\circ}_{rxn}= -151kJ                                

(a) If \Delta H_{o}^{f}of NaOCl(aq)  = -346 kJ/mol, find of N_2H_4(aq)

(b) What is the heat released when aqueous  is added to 5.00 x 103 L of plant water that is 2.50 x 104 M ?

                                                         Step by step solution

Step 1 of 2

(a)

The thermochemical equation is as follows:

.......(1)

Use the following equation to determine the standard heat of formation  for  from equation ( 1 ).

                   

Here  represents the sum of terms, m and n are the number of moles of products and reactants, respectively. The term  is the standard heat of formation.

=34 kJ/mol

Thus, the standard heat of formation  for  will be 34 kJ / mol.

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