The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
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Answer:
d) V = 91.3 L
Explanation:
Given data:
Volume of nitrogen = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Number of atoms of nitrogen = 2.454×10²⁴ atoms
Solution:
First of all we will calculate the number of moles of nitrogen by using Avogadro number.
1 mole = 6.022×10²³ atoms
2.454×10²⁴ atoms × 1 mol / 6.022×10²³ atoms
0.407×10¹ mol
4.07 mol
Volume of nitrogen:
PV = nRT
1 atm × V = 4.07 mol ×0.0821 atm.L /mol.K ×273.15 K
V = 91.3 atm.L /1 atm
V = 91.3 L
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
Answer:
<h2>0.93 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 37.2 g
volume = 40 mL
We have
We have the final answer as
<h3>0.93 g/mL</h3>
Hope this helps you
