更新1:
I just realized that in a rush I thought the arrow was after clo2, therefore accidentally calculating clo2 as a reactant. thanks for the responses guys
Basic Enthalpy problem, need help ASAP!?
2017-08-18 10:14 am
so when I solve for enthalpy with (d is delta) dH=[dHreactants]-[dHproducts] or even dH=[dHproducts]-[dHreactants] I always get roughly 1850 ( I rounded the enthalpies to -350, -400, and -250) I honestly cant figure out their reasoning on how the enthalpy is near 3000, my mcat is in 2 days so a quick response would be appreciated! thank you
回答 (3)
2017-08-18 10:35 am
Thermochemical equation for the decomposition of "2 moles" of nitroglycerin at 25°C:
2C₃H₅N₃O₉(l) → 6CO₂(g) + 5H₂O(g) + 3N₂(g) + (1/2)O₂(g) …. ΔH° = ?
ΔH° = ∑ΔHf°[products] - ∑ΔHf°[reactants]
= 6ΔHf°[CO₂(g)] + 5ΔHf°[H₂O(g)] + 3ΔHf°[N₂(g)] + (1/2)ΔHf°[O₂(g)] - 2ΔHf°[H₅N₃O₉(l)]
≈ 6×(-400) + 5×(-250) + 3×(0.0) + (1/2)×(0.0) - 2×(-350) kJ
= -2950 kJ
The closest option: D. 3000
2C₃H₅N₃O₉(l) → 6CO₂(g) + 5H₂O(g) + 3N₂(g) + (1/2)O₂(g) …. ΔH° = ?
ΔH° = ∑ΔHf°[products] - ∑ΔHf°[reactants]
= 6ΔHf°[CO₂(g)] + 5ΔHf°[H₂O(g)] + 3ΔHf°[N₂(g)] + (1/2)ΔHf°[O₂(g)] - 2ΔHf°[H₅N₃O₉(l)]
≈ 6×(-400) + 5×(-250) + 3×(0.0) + (1/2)×(0.0) - 2×(-350) kJ
= -2950 kJ
The closest option: D. 3000
2017-08-18 10:39 am
There is no reason to round the enthalpies. Use them as they are given.
products - reactants:
12 x (- 393.5) + 10 x (- 241.8) + 6 x (0.0) + 1 (0.0) - 4 x (- 364.0) = - 5684 kJ
The amount just calculated is for 4 moles of nitroglycerin, but the question asks about 2 moles, so:
(- 5684 kJ) x (2 mol / 4 mol) = - 2842 kJ
This is closer to answer D. than to any of the other proposed answers -- although I see no reason to round the answer to just one significant digit.
products - reactants:
12 x (- 393.5) + 10 x (- 241.8) + 6 x (0.0) + 1 (0.0) - 4 x (- 364.0) = - 5684 kJ
The amount just calculated is for 4 moles of nitroglycerin, but the question asks about 2 moles, so:
(- 5684 kJ) x (2 mol / 4 mol) = - 2842 kJ
This is closer to answer D. than to any of the other proposed answers -- although I see no reason to round the answer to just one significant digit.
2017-08-18 10:28 am
The problem says react 2 moles of nitroglycerin. The equation gives 4 moles starting material, so divide the mole amounts by 2 to see what the reaction would yield with just 2 moles. According to the equation, then, 2 moles nitroglycerin would give 6 moles CO2 and 5 moles H2O.
So now multiplying the mole amounts by the enthalpies and setting up products - reactants:
(5 moles H2O)(-241.8 KJ/mol H2O) + (6 moles CO2)(-393.5) - (2moles C3H5N3O9)(-364KJ/mol C3H5N3O9) which should give -2842 KJ.
So now multiplying the mole amounts by the enthalpies and setting up products - reactants:
(5 moles H2O)(-241.8 KJ/mol H2O) + (6 moles CO2)(-393.5) - (2moles C3H5N3O9)(-364KJ/mol C3H5N3O9) which should give -2842 KJ.
收錄日期: 2021-05-01 15:00:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170818021459AAdgjQG