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Question: Bond energies of \(N \equiv N\) ,\(H - H\),\(N - H\) bonds are \(945\),\(463\) & \(391KJmo{l^{ - 1}}...

Bond energies of N≡NN \equiv N ,H−HH - H,N−HN - H bonds are 945945,463463 & 391KJmol−1391KJmo{l^{ - 1}} respectively, the enthalpy of the following reaction is,
N2(g)+3H2(g)→2NH3(g){N_2}\left( g \right) + 3{H_2}\left( g \right)\xrightarrow{{}}2N{H_3}\left( g \right)

Explanation

Solution

We know that the change of entropy for phase transition of substance is adequate to the ratio of change of enthalpy and therefore the temperature at which the change is happening. This is often into consideration that the encompassing pressure remains constant during the phase transition.

Complete step by step answer:
First, we see what's enthalpy of a reaction is.
Enthalpy: The heat of reaction or the enthalpy change is defined because the energy absorbed or released in any reaction.
The overall reaction enthalpy change is certainly obtained via Hess' Law:
ΔHrxn=ΔH1+ΔH2\Delta {H_{rxn}} = \Delta {H_1} + \Delta {H_2}
The given reaction is,
N2(g)+3H2(g)→2NH3(g){N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)
Given data,
The bond energy of N≡NN \equiv N is 945945
The bond energy of N−HN - H is 391391
The bond energy of H−HH - H is 463463.
We can calculate the enthalpy of the reaction by using the below formula as,
Enthalpy of the reaction = bond energies of the reactants – bond energies of products.
In the given reaction, reactants are H−HH - Hand N≡NN \equiv N and the product of this reaction is ammonia the bond between them is N−HN - H.
Enthalpy of the reaction =3×(H−H)+(N≡N)−2×3×(N−H) = 3 \times \left( {H - H} \right) + \left( {N \equiv N} \right) - 2 \times 3 \times \left( {N - H} \right)
Now substitute the values of bond energy we get,
Enthalpy of the reaction =3×(436)+(945)−6×(391) = 3 \times \left( {436} \right) + \left( {945} \right) - 6 \times \left( {391} \right)
Enthalpy of the reaction =1308+945−2346=2253−2346 = 1308 + 945 - 2346 = 2253 - 2346
On simplification we get,
Enthalpy of the reaction =−93KJ = - 93KJ
Thus the enthalpy of the reaction is −93KJ - 93KJ.

Additional Information:
Enthalpy of fusion:
The heat which a solid absorbs when it melts is termed the enthalpy of fusion or heat of fusion and is usually quoted on a molar basis. as an example , when of ice is melted, we discover the molar enthalpy of fusion of ice is , and that we can write that as,
H2O(s)→0∘CH2O(l){H_2}O\left( s \right)\xrightarrow{{{0^ \circ }C}}{H_2}O\left( l \right)
ΔHm=6.01KJ/mol\Delta {H_m} = 6.01KJ/mol
Enthalpy of Vaporization:
When a liquid is boiled, the difference of temperature with the warmth supplied is analogous thereto found for melting. When heat is supplied at a mild rate to a liquid at atmospheric pressure, the temperature rises until the boiling point is attained. After the boiling point the temperature remains constant until the enthalpy of vaporization is supplied and again the temperature rises once all the liquid has been converted to vapor.

Note:
If energy is absorbed by the reaction it is claimed to be endothermic and is positive and more energy is required to interrupt the bonds and therefore the reactants are lower in energy than the products.
If energy is free by the reaction is claimed to be exothermic and is negative and more energy is released while forming bonds and therefore the energy of the merchandise is less than the reactants.