Respuesta :
Answer:
∆H° = −151.2 kJ/mol.
Explanation:
Applying Hess law:
Overall equation,
N2H4(ℓ) + H2(g) → 2NH3(g)
i. N2H4(ℓ) + O2(g) → N2(g) + 2H2O(g) ∆H° = −543 kJ/mol
Flipping,
ii. 2H2(g) + O2(g) → 2H20(g)
to this:
ii. 2H20(g) → 2H2(g) + O2(g)
∆H° = +484kJ/mol
iii. N2(g) + 3 H2(g) → 2 NH3(g)
∆H° = −92.2 kJ/mol
Cancelling reactants on opposite sides,
N2H4(ℓ) + H2(g) → 2 NH3(g)
Calculating Enthalpy, by adding all the enthalpies
(−543 kJ/mol) + (+484kJ/mol) + (−92.2 kJ/mol)
∆H° = −151.2 kJ/mol.
The standard enthalpy for the reaction of hydrazine has been -151.2 kJ/mol.
The standard enthalpy of the reaction for the formulation of the product in the balanced chemical reaction. The reaction has been given as:
[tex]\rm N_2H_4\;+\;H_2\;\rightarrow\;2\;NH_3[/tex]
The reaction has been achieved by:
[tex]\rm N_2H_4\;+\;O_2\;\rightarrow\;N_2\;+\;2\;H_2O[/tex] = -543 kJ/mol
[tex]\rm 2\;H_2\;+O_2\;\rightarrow\;H_2O[/tex] = -484 kJ/mol
Flipping the equation for the dissociation of a water molecule:
[tex]\rm H_2O\;\rightarrow\;2\;H_2\;+\;O_2[/tex] = 484 kJ/mol
The formed hydrogen has been reacted with the nitrogen for the formation of ammonia.
[tex]\rm N_2\;+3\;H_3\;\rightarrow\;2\;NH_3[/tex] = -92.2 kJ/mol
The summation of reactions has been:
[tex]\rm N_2H_4\;+O_2\;+\;H_2O\;+\;N_2\;+\;3\;H_3\;\rightarrow\;N_2\;+2\;H_2O\;+\;2\;H_2\;+\;O_2\;+\;2\;NH_3[/tex]
Subtracting the compounds that have been common for both the products and the reactants, the reaction has been:
[tex]\rm N_2H_4\;+\;H_2\;\rightarrow\;2\;NH_3[/tex]
Thus, the reaction enthalpy has been given by:
Standard enthalpy of the reaction = -543 + 484 + -92.2 kJ/mol
Standard enthalpy of the reaction = -151.2 kJ/mol
The standard enthalpy for the reaction of hydrazine has been -151.2 kJ/mol.
For more information about the standard enthalpy, refer to the link:
https://brainly.com/question/10583725
