Given... #N_2(g) + 3H_2(g) rarr 2NH_3(g) + 92.4*kJ#... ..what is #DeltaH_f^@# #NH_3(g)#?

Answer 1

#DeltaH_f^@# #NH_3(g)# #=# #-46.2*kJ*mol^-1#.

By definition, #DeltaH_f^@# is the enthalpy change associated with the formation of 1 mole of product from its constituent elements in their standard states at #298*K#.

I think you have quoted the enthalpy change for the reaction:

#N_2(g) + 3H_2(g) rarr 2NH_3(g)#, which of course is TWICE the enthalpy of formation for #NH_3#, because this reaction formed TWO moles of ammonia.

Finally, we can write:

#1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)# #;DeltaH_"rxn"^@=DeltaH_f^@"ammonia"=-46.2*kJ*mol^-1#.
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Answer 2

To find the standard enthalpy of formation (ΔH_f°) of NH₃(g), we can use Hess's Law.

Given: N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -92.4 kJ

Since the enthalpy change for the reaction is given as -92.4 kJ, we can infer that it is the enthalpy change for the formation of 2 moles of NH₃(g) from its elements in their standard states. Therefore, the standard enthalpy of formation of NH₃(g) is -92.4 kJ / 2 = -46.2 kJ.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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