given that: 2NO (g) + O2 (g) ----> 2NO2 (g) calculate delta H^0 for the followin

Question

given that: 2NO (g) + O2 (g) ----> 2NO2 (g) calculate delta H^0 for the following reaction: No2 (g) -----> NO (g) + 1/2 O2 (g) Please show all work

Explanation / Answer

The Hess law is used to discover the reaction's entalphy and doing it, you must cancel the same substances, if they are in differents sides, to make the given reaction. Remember that if you multiply or divide the reaction, the same happens with the entalphy and when you invert the reaction, the entalphy's sign must be also inverted. The first one is easy, but the second one took me some work. 1) The reaction you have is: 2 NO + O2 -> 2 NO2 It means that 2 NO and O2 must be on the left side and 2 NO2, on the right one, so you will look for the entalphies that were given and invert and/or multiply to copy this order, look: 1/2 N2(g) + O2(g) --> NO2(g), deltaH = 33.2 kJ (should be multiplied by 2) N2 + 2 O2 -> 2 NO2.........deltaH = +66,4 kJ 1/2 N2(g) + 1/2O2(g) --> NO(g), delta H = 90.2 kJ (should be multiplied by 2 and inverted) 2 NO -> N2 + O2..............deltaH = -180,4 kJ Now you have: N2 + 2 O2 -> 2 NO2.........deltaH = +66,4 kJ 2 NO -> N2 + O2..............deltaH = -180,4 kJ --------------------------------------… canceling both N2 and 2 O2 with O2, you will find the given reaction: 2 NO + O2 -> 2 NO2........deltaH = 66,4 - 180,4 = -114,0 kJ

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

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