# Given the ammonia synthesis: N_2(g) + 3H_2(g) rarr 2NH_3(g) DeltaH_"rxn"^@=-92.6*kJ*mol^-1, what reaction enthalpy is associated with the formation of 8.55xx10^4*g of ammonia?

Jun 30, 2017

How? We use the enthalpy as a variable in the reaction..........and get $\Delta {H}_{\text{rxn}} = - 232725 \cdot k J \ldots \ldots . .$

#### Explanation:

We got...

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \rightarrow 2 N {H}_{3} \left(g\right) + 92.6 \cdot k J$

Which tells that the formation of $34 \cdot g$ ammonia releases $2 \times 92.6 \cdot k J$. Agreed?

But here we have made much more than 2 mol ammonia, and the enthalpy changes appropriately.

$\text{Moles of ammonia} = \frac{8.55 \times {10}^{4} \cdot g}{17.01 \cdot g \cdot m o {l}^{-} 1} = 5026.5 \cdot m o l$.

And thus $\Delta {H}_{\text{rxn}} = \frac{- 5026.5 \cdot m o l \times 92.6 \cdot k J \cdot m o {l}^{-} 1}{2}$

=??*kJ

Why is $\Delta {H}_{\text{rxn}}$ negative?