What is Delta H degrees for the gas-phase reaction below?
CH4+ CL3 = CH3CL+ HCL
Bond dissociation energies of CL-CL, C-CL, H-CL, and H-C are 243, 339, 431, and 414kj/mol respectively
a)113kJ
b)-113kJ
c)263kJ
d)-263kJ
Can somebody teach me how to solve this?
B.E. = bond energy.
delta Hrxn = (sum B.E.reactants)-(sum B.E. products)
DErxn = [DE(C-H) + DE(Cl-Cl)] -[DE(C-Cl) + DE(H-Cl)] = ??
If I didn't goof the value is -113 kJ/mol but check me out on that.This procedure is good ONLY for estimating delta Hrxn and it is good for gas phase reactions only. The better value is obtained by looking up delta H formations and DErxn = (sum products)-(sum reactants) [Note the difference between the two procedures. From BE it is (sum reactants)- (sum products) and from DHformation it is (sum products) - (sum reactants).
To solve this problem, we need to calculate the ΔH degrees for the given gas-phase reaction.
ΔH degrees represents the change in enthalpy (heat) during a chemical reaction at standard conditions (typically 25°C and 1 atm). It can be calculated using the following formula:
ΔH degrees = Σ (ΔH bonds broken) - Σ (ΔH bonds formed)
Step 1: Identify the bonds broken and formed in the equation:
Bonds broken:
- CL-CL: 1
- C-CL: 1
Bonds formed:
- H-CL: 1
- H-C: 1
(Note: The numbers next to the bonds indicate the stoichiometric coefficients in the balanced equation.)
Step 2: Look up the bond dissociation energies for the bonds involved:
- CL-CL: 243 kJ/mol
- C-CL: 339 kJ/mol
- H-CL: 431 kJ/mol
- H-C: 414 kJ/mol
Step 3: Calculate the bond energy changes:
ΔH bonds broken = (1 x 243 kJ/mol) + (1 x 339 kJ/mol) = 582 kJ/mol
ΔH bonds formed = (1 x 431 kJ/mol) + (1 x 414 kJ/mol) = 845 kJ/mol
Step 4: Calculate the ΔH degrees:
ΔH degrees = Σ (ΔH bonds broken) - Σ (ΔH bonds formed)
ΔH degrees = 582 kJ/mol - 845 kJ/mol = -263 kJ/mol
Therefore, the answer is (d) -263 kJ.