Achiever gchem test 2

[CSU Undergrad]

---

Members do not see this ad.

heres the question

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

This is what I get:

-312.27 and here is how I set it up

[3(565) + 3(N-F)] - [3(389) + 3(155)] = -873.81

But the answer key says 270 kJ, what am I doing wrong?

 

[jntruong2003]

the formula is reactants (breaking bonds) - products (bond formation) = enthalpy change

you have products - reactants.

if you calculate w/ the formula above, you should get -270.27.

hope that helps

 

[CSU Undergrad]

ahh thank you.

 

[prydA]

just remember that energy is required in order to break bonds and energy is released when bonds are formed.. which leads to the equation reactants (breaking bonds) - products (bond formation) = enthalpy change.

at least that's how i think in order to remember that equation.

 

[rose786]

I could have sworn one of my books said enthalpy= sum of products - reactants. No wonder I always get enthalpy problems wrong.