Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ/mole) the standard enthalpy change DH° for the reaction written below, using the bond energies given. Just enter a number (no units).
N2(g) + 3H2(g) -> 2NH3(g)Bond: N≡N H-H N-HBond energy: 945 432 391

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kobenhavn kobenhavn · Answer 1 · 2019-12-18T05:50:08+01:00

Answer: -105 kJ

Explanation:

The balanced chemical reaction is,

[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)][/tex]

[tex]\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})][/tex]

[tex]\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})][/tex]

where,

n = number of moles

Now put all the given values in this expression, we get

[tex]\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)][/tex]

[tex]\Delta H=-105kJ[/tex]

Therefore, the enthalpy change for this reaction is, -105 kJ

adioabiola adioabiola · Answer 2 · 2021-11-04T05:39:41+01:00

The standard enthalpy change DH° for the reaction between Nitrogen and hydrogen to form ammonia is; DH° = 105 kJ

According to the question;

The balanced chemical reaction for the reaction of Nitrogen, N2 and Hydrogen molecule, H2 to yield Ammonia, NH3 is as follows;

It is evident from the chemical equation that;

In essence, the enthalpy change is given as;