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

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Answer 1 · 2017-06-30T04:37:59

How? We use the enthalpy as a variable in the reaction..........and get $DeltaH_"rxn"=-232725*kJ........$

We got...

$N_2(g) + 3H_2(g) rarr 2NH_3(g)+92.6*kJ$

Which tells that the formation of $34*g$ ammonia releases $2xx92.6*kJ$ . Agreed?

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

$"Moles of ammonia"=(8.55xx10^4*g)/(17.01*g*mol^-1)=5026.5*mol$ .

And thus $DeltaH_"rxn"=(-5026.5*molxx92.6*kJ*mol^-1)/2$

$=??*kJ$

Why is $DeltaH_"rxn"$ negative?

Given the ammonia synthesis: N_2(g) + 3H_2(g) rarr 2NH_3(g)N_2(g) + 3H_2(g) rarr 2NH_3(g) DeltaH_"rxn"^@=-92.6*kJ*mol^-1DeltaH_"rxn"^@=-92.6*kJ*mol^-1, what reaction enthalpy is associated with the formation of 8.55xx10^4*g8.55xx10^4*g of ammonia?