Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: armim003 on February 15, 2020, 05:00:24 AM

There's a chemistry problem I'm trying to solve. Burning 1 kg of syntin or 1 kg of hydrogen will release more E? We predict that syntin decomposes into H2O and CO2. Their standard ΔH of formation: ΔH (CO2, g) =393,5kJ/mol; ΔH(H2O) =241,8 kJ/mol. Many thanks! :)

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Oh, thank you. :) I will do so.

The usual combustion of Syntin is with pure oxygen, and despite high pressure, the high temperature makes about as much CO as CO_{2}.
Anyway, you can compute a heat of combustion, or you can answer qualitatively that the same mass of hydrogen releases more combustion heat than a hydrocarbon, because it contains many more moles. I know no exception. The margin is big.
"Energy" isn't accurate enough, and "E" seems to designate the internal energy in some countries. You're probably talking about the enthalpy H.

Thank you, sir. :) So, I thought about a possible way how to calculate it... Here it is.
a) Burning 1 kg of H2.
2 H2 (g) + O2 (g) => 2 H2O (g)
Mr = 2
Q1 = 2 .  241,8 =  483,6 kJ.mol1
n = m/M = 1000 g / 2 g.mol1 = 500 mol
E = 500 .  483,6 =  241 800 kJ =  241,8 MJ
b) Burning 1 kg of syntin.
C10H16 + 14 O2 => 10 CO2 + 8 H2O
Mr = 136,238
Q2 = (10 . – 393,5) + (8 .  241,8) =  5889,4 kJ.mol1
n = m/M = 1000/136,238 = 7,34 mol
E = 7,34 .  5889,4 =  43228,196 kJ =  43,228196 MJ.
Is that right? :D I presume not, but otherwise I cannot come to any other solution. Thanks.

In a, what you have shown is the burning of 2 moles of H_{2}. Therefore the enthalpy of combustion is 241.8 kJ/mol, and E = 120.9 MJ.
The answer for b looks correct, but it has far too many sig figs.

I don't see the enthalpy of formation of Syntin anywhere in the enthalpy of combustion. It should be there. +133kJ/mol (Syntin is nicely endothermic).
(The enthalpy H is an energy. It's the one used in such computations. "Energy" isn't accurate enough. But for high school, this difference is normally ignored.)
500 moles H_{2} produce 500 moles H_{2}O. 241.8kJ/mol is the enthalpy of formation for gaseous water at 298K. You compute here the lower calorific power of the fuel. Condensed water releases more heat, which defines the higher calorific power. And at a different temperature, the result would differ too.
Syntin is a rocket fuel, hence burned with oxygen. At that price, I don't imagine torching it in a stove. But in a rocket, the hot combustion would produce much CO, so computing with CO_{2} is quite artificial.
We can guess your symbols Q, n, m... but they are no universal convention.