Given... #N_2(g) + 3H_2(g) rarr 2NH_3(g) + 92.4*kJ#... ..what is #DeltaH_f^@# #NH_3(g)#?
I think you have quoted the enthalpy change for the reaction:
Finally, we can write:
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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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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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