Which of the following reactions will have delta H < 0 and delta S > 0? I think this means exothermic and increase in entropy.
1) O3 (g) --> O2 (g) + O
2) 5 C (s) + 4 KNO3 (s) --> 5 CO2 (g) + 2N2 (g) + 2 K2O (s)
3) Neither
It's on a practice exam, and answers are not given...
Thanks!

Okay, so Delta H is negative so its an endothermic rxn, and Delta S is postiive so an increase in entropy. 1) Your Delta n of gas is bigger then 1 so it is increasing entropy BUT you are breaking 3 bonds O-O=O (requires E, so endothermic) to make 2 bonds O=O (Release E) so its its an overall endothermic rxn, its not 1)

2) You are making gas, theres your increase in entropy, and you are making more bonds then you are breaking, so more energy is released then is required thus its exothermic. so 2) is correct!

Best of luck on the exam!

Originally Posted by philkolb31

Okay, so Delta H is negative so its an endothermic rxn, and Delta S is postiive so an increase in entropy. 1) Your Delta n of gas is bigger then 1 so it is increasing entropy BUT you are breaking 3 bonds O-O=O (requires E, so endothermic) to make 2 bonds O=O (Release E) so its its an overall endothermic rxn, its not 1)

2) You are making gas, theres your increase in entropy, and you are making more bonds then you are breaking, so more energy is released then is required thus its exothermic. so 2) is correct!

Best of luck on the exam!

Hey, correct me if I wrong but I think delta H negative is exothermic and O3 => O2 + O delta n(gas) is +1 so entropy increased. According to Gibbs energy, delta G = delta H - T(delta S) = - -(+) = - which means spontaneous and exothermic at any T.

Oh yep my bad - delta H means its exothermic