Question

hydrazine (n₂h₄) is a fuel used by some spacecraft. it is normally oxidized by n₂o₄ according to the equation: n₂h₄(l)+n₂o₄(l)→2n₂o(g)+2h₂o(g)
complete the following table of standard enthalpies of formation:

compound	δh°f (kj/mol)
n₂o₄(l)	9.16
n₂o(g)	82.18
h₂o(g)	-241.8

• part a

calculate δrh° for this reaction using standard enthalpies of formation.
express your answer in kilojoules per mole to one - decimal place.
δrh° =
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Explanation:

Step1: Recall the formula for $\Delta H^{\circ}$

$\Delta H^{\circ}=\sum n_p\Delta H_f^{\circ}(products)-\sum n_r\Delta H_f^{\circ}(reactants)$

Step2: Identify products and reactants and their coefficients

The reaction is $N_2H_4(l)+N_2O_4(l)
ightarrow 2N_2(g)+2H_2O(g)$. For products: $n_{N_2}=2$, $\Delta H_f^{\circ}(N_2(g)) = 0$ (by definition for elements in standard state), $n_{H_2O}=2$, $\Delta H_f^{\circ}(H_2O(g))=-241.8\ kJ/mol$. For reactants: $n_{N_2H_4}=1$, $\Delta H_f^{\circ}(N_2H_4(l)) = 50.6\ kJ/mol$, $n_{N_2O_4}=1$, $\Delta H_f^{\circ}(N_2O_4(l)) = -19.5\ kJ/mol$.

Step3: Calculate $\sum n_p\Delta H_f^{\circ}(products)$

$\sum n_p\Delta H_f^{\circ}(products)=2\times0 + 2\times(-241.8)=-483.6\ kJ/mol$

Step4: Calculate $\sum n_r\Delta H_f^{\circ}(reactants)$

$\sum n_r\Delta H_f^{\circ}(reactants)=1\times50.6+1\times(-19.5)=50.6 - 19.5=31.1\ kJ/mol$

Step5: Calculate $\Delta H^{\circ}$

$\Delta H^{\circ}=-483.6 - 31.1=-514.7\ kJ/mol$

Answer:

$-514.7$