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:
4.33 L
Explanation:
Step 1: Given data
Initial volume of the balloon (V₁): 3.00 L
Initial pressure of the balloon (P₁): 765 torr
Final volume of the balloon (V₂): ?
Final pressure of the balloon (P₂): 530 torr
Step 2: Calculate the final volume of the balloon
If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.
Answer:heat-,7
Explanation:According to table P, heat- is an organic prefix used to represent 7 carbon atoms
