chem ????????

[Denro]

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Given the average bond energies, N-H(389 kJ/mol), H-F(565 kJ/mol), and F-F(155 kJ/mol), calculate the average bond energy for N-F.
NH3(g) + 3F2(g) à NF3(g) + 3HF(g) ∆H = -873.81 kJ




A. 936.81 kJ/mol
B. 810.81 kJ/mol
C. 312.27 kJ/mol
D. 270.27 kJ/mol
E. 63 kJ/mol


ans D

Enthalpy change,
this question is from the DAT achiever and ive never seen negative G in there.
i thought it was always H= product - reactant

∆H = -G ∆H(product bonds) + G ∆H(reactant bonds)

 

[dlink]

I believe this must have been due to character/symbol mismatch that often occurs on most computers. Mine shows up as the Greek summation symbol like the capitalized "E" instead of "G" in the solution page.

Is there any other part(s) from this question you've been confused with other than this minor issue?

 

[rals]

Given the average bond energies, N-H(389 kJ/mol), H-F(565 kJ/mol), and F-F(155 kJ/mol), calculate the average bond energy for N-F.
NH3(g) + 3F2(g) à NF3(g) + 3HF(g) ∆H = -873.81 kJ




A. 936.81 kJ/mol
B. 810.81 kJ/mol
C. 312.27 kJ/mol
D. 270.27 kJ/mol
E. 63 kJ/mol


ans D

the answer should be C according to my calculations.... I came up with exactly 312.27... correct me if i'm wrong

 

[dlink]

the answer should be C according to my calculations.... I came up with exactly 312.27... correct me if i'm wrong

How did you arrive at C?

 

[groovier]

the answer should be C according to my calculations.... I came up with exactly 312.27... correct me if i'm wrong

This is a typical exothermic reaction.

3*389+3*155+873.81=3*x+3*565
3x=810.81
x=270.27 (KJ/mol)

Total energy absorbed to break bonds - Total energy released to form bonds = (3*389+3*155) - (3*270.27+3*565) = -873.81 kJ/mol.
Thus, in an exothermic reaction, 873.81 kJ of energy are released to the
surroundings.

 

[rals]

ok, i got it... i just mistakenly did the bonds formed - bonds broken.

thanks for correcting me!