Respuesta :
Answer:
The enthalpy of the reaction is coming out to be -380.16 kJ.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
For the given chemical reaction:
[tex]N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})][/tex]
We are given:
[tex]\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo[/tex]
Putting values in above equation, we get:
[tex]\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ[/tex]
Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.
dH∘rxn for this reaction using standard enthalpies of formation is mathematically given as
dH_{rxn}=-380.16 kJ
What is dH∘rxn for this reaction using standard enthalpies of formation.?
Question Parameter(s):
Hydrazine (N2H4) is a fuel used by some spacecraft.giving
N2H4(l)+N2O4(g)→2N2O(g)+2H2O(g)
Generally, the equation for the is mathematically given as
[tex]d H_{rxn}=\sum [n* d H_f(product)]-\sum [n*d H_f(reactant)][/tex]
Therefore
[tex]d H_{rxn}=[(2* 81.6 )+2* -241.8 )]-[(1 * (50.6 ))+(1 * (9.16))[/tex]
dH_{rxn}=-380.16 kJ
In conclusion, dH∘rxn for this reactionis
dH_{rxn}=-380.16 kJ
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