[slu1986]

When 0.215 mol NH3 is formed from nitrogen and hydrogen, 9.91 kJ of heat is released as heat. Assuming constant pressure is maintained, what is the ΔH for this reaction per mole of NH3 formed?

A) 46.1 kJ B) 9.91 kJ C) -9.91 kJ D) -2.13 kJ E) -46.1 kJ

This is how I worked the problem out, but I'm not sure if its the right way to solve it.

0.215 mol NH3 * 9.91 kJ/1 mol NH3 = -2.13 <---- I put the answer as negative b/c heat is being released to the surroundings. i.e. heat is lost from the system. 

but, then I'm not sure if you work it this way= 9.91 kJ/ 0.215 mol = -46.1 kJ

I'm just confused about how to approach solving this.  If you could please help me, I would greatly appreciate it. Thanks     

[Dan]

but, then I'm not sure if you work it this way= 9.91 kJ/ 0.215 mol = -46.1 kJ

This is numerically correct, but watch your units:

Δ_mH = 9.91 kJ/0.215 mol = -46.1 kJ/mol (or kJ mol^-1)

This is the molar (this is what the subscript m denotes) enthalpy change, so when 1 mol ammonia is formed, 46.1 kJ of heat is released:

ΔH = Δ_mH x (moles of ammonia)

so for 1 mol of ammonia being formed:

ΔH = -46.1 kJ mol^-1 x 1 mol = -46 kJ

I'm probably being too picky here.

Conceptually, you can consider that the formation 0.2 mol of ammonia releases about 10 kJ, so you should expect the formation of 1 mol to release about five times as much heat, since you are forming about five times as much ammonia - so you should expect to get an answer around -50 kJ - and as you say ΔH is negative since heat is released in this reaction.

[slu1986]

So are you saying that I solved the problem right the second way that I worked it to get -46.1 kJ?

[Dan]

yes

[slu1986]

Thank you so much for your help..I really appreciate it.