Answer H° f of N2O5(g) = +11.3kJ/mol Explanation This problem is solved by Hess\

Question

Answer

H°f of N2O5(g) = +11.3kJ/mol

Explanation

This problem is solved by Hess's Law of Constant Heat of Summation

2H2(g) + O2(g) -------> 2H2O(l) H= -571.6kJ ------ eqn1

N2O5(g) + H2O(l) -------> 2HNO3(l) H = -73.7kJ ------ eqn2

1/2N2(g) + 5/2O2(g) + 1/2H2(g) ------> HNO3(l) H = -174.1kJ -------eqn3

reverse eqn2 and multiply it by 2

4HNO3(l) -------> 2N2O5(g) + 2H2O(l) H= +147.4kJ -------eqn4

multiply eqn3 by 4

2N2(g) + 6O2(g) + 2H2(g) ------> 4HNO3(l) H= -696.4kJ -------eqn5

reverse eqn 1

2H2O(l) -------> 2H2(g) + O2(g) H = +571.6kJ -----eqn6

add eqn 4 , 5 and 6

4HNO3(l) + 2N2(g) + 6O2(g) + 2H2(g) + 2H2O(l) -------> 2N2O5(g) + 2H2O(l) +4HNO3 + 2H2(g) + O2(g)

After cancallisation

2N2(g) + 5O2(g) ------ > 2N2O5(g) H= +22.6kJ -----eqn7

divide eqn7 by 2

N2(g) + 5/2O2(g) ------->N2O5(g) H= +11.3kJ

Explanation / Answer

GO-2 Thermochemistry, Calorimetry, Heats of Reaction 11. Given the following thermochemical equations A 2H2(8) +02l8)2H-o(e) B N2Os(g) + H20(e)> 2HNO(e) H =-571.6 kJ H =-73.7 kJ Calculate H for the formation of one mole of dinitrogen pentoxide from its elements in their standard state at 25 °C and 1 atm. (15 pts.)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.