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:
The answer to your question is AgCl
Explanation:
Data
Silver = Ag = 75%
Chlorine = Cl = 25%
1.- Convert percent numbers to grams
Silver = 75 g
Chlorine = 25 g
2.- Calculate the moles of each element
Mass number Ag = 108 g
Mass number Cl = 35.5 g
108 g of Ag ------------------ 1 mol
75 g of Ag ------------------ x
x = (75 x 1) / 108
x = 75 / 108
x = 0.69 moles
35.5 g of Cl ------------------ 1 mol
25 g of Cl ----------------- x
x = (25 x 1) / 35.5
x = 0.70 moles
3.- Divide by the lowest number of moles
Ag = 0.69/0.69 = 1
Cl = 0.70 / 0.69 = 1.02 ≈ 1
4.- Write the empirical formula
Ag₁ Cl₁ = AgCl
