Given, \(N_2(g) + 3H_2(g)→2NH_3(g)\); \(∆_rH^Θ=-92.4\ kJ mol^{–1}\)
What is the standard enthalpy of formation of \(NH_3\) gas?
What is the standard enthalpy of formation of \(NH_3\) gas?
Solution and Explanation
Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.
Re-writing the given equation for 1 mole of \(NH_3(g)\),
\(\frac 12 N_2(g) + \frac 32H_2(g) → NH_3(g)\)
Standard enthalpy of formation of \(NH_3\)
= \(\frac 12 ∆_rH^Θ\)
= \(\frac 12(–92.4\ kJ mol^{–1})\)
= \(–46.2\ kJ mol^{–1}\)
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Concepts Used:
Enthalpy change
Enthalpy Change refers to the difference between the heat content of the initial and final state of the reaction. Change in enthalpy can prove to be of great importance to find whether the reaction is exothermic or endothermic.
Formula for change in enthalpy is:-
dH = dU + d(PV)
The above equation can be written in the terms of initial and final states of the system which is defined below:
UF – UI = qP –p(VF – VI)
Or qP = (UF + pVF) – (UI + pVI)
Enthalpy (H) can be written as H= U + PV. Putting the value in the above equation, we obtained:
qP = HF – HI = ∆H
Hence, change in enthalpy ∆H = qP, referred to as the heat consumed at a constant pressure by the system. At constant pressure, we can also write,
∆H = ∆U + p∆V
Standard Enthalpy of Reaction
To specify the standard enthalpy of any reaction, it is calculated when all the components participating in the reaction i.e., the reactants and the products are in their standard form. Therefore the standard enthalpy of reaction is the enthalpy change that occurs in a system when a matter is transformed by a chemical reaction under standard conditions.