Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Answer: A) Water Explanation: Water molecules are polar molecules
Answer:
-125 kJ
Explanation:
You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.
H₂C=CH₂ + H₂ ⟶ H₃C-CH₃
Bonds: 4C-H + 1C=C 1H-H 6C-H + 1C-C
D/kJ·mol⁻¹: 413 612 436 413 347
The formula relating ΔHrxn and bond dissociation energies (D) is
ΔHrxn = Σ(Dreactants) – Σ(Dproducts)
(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)
<em>For the reactant</em>s:
Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ
<em>For the products:</em>
Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ
<em>For the system</em>
:
ΔHrxn = 2700 - 2825 = -125 kJ
Hi,
Answer is 191.2.
800J = 191.2 cal
Hope this helps.