Question

Hydrazine (N2H4) is a fuel used by some spacecraft. It is normally oxidized by N2O4 according to the following equation: N2H4(l)+N2O4(g)→2N2O(g)+2H2O(g) Part A Calculate ΔH∘rxn for this reaction using standard enthalpies of formation. Express your answer using four significant figures.

Answer 1

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 $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]$

We are given:

$\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ$

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.