Question

hydrazine (n2h4) is a fuel used by some spacecraft. it is normally oxidized by n2o4 according to the equation: n2h4(l)+n2o4(g)→2no2(g)+2h2o(l) consider the following table of standard enthalpies of formation: compound δfh° (kj mol−1) n2h4(l) 50.6 n2h4(g) 95.4 no2(g) 33.2 n2o4(g) 9.16 n2o(g) 81.6 h2o(l) −285.8 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.

Explanation:

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

$\Delta_{r}H^{\circ}=\sum n_{products}\Delta_{f}H^{\circ}_{products}-\sum n_{reactants}\Delta_{f}H^{\circ}_{reactants}$

Step2: Identify products and reactants and their coefficients

For the reaction $N_{2}H_{4}(l)+N_{2}O_{4}(g)
ightarrow 2NO_{2}(g)+2H_{2}O(l)$, the products are $2$ moles of $NO_{2}(g)$ and $2$ moles of $H_{2}O(l)$, and the reactants are $1$ mole of $N_{2}H_{4}(l)$ and $1$ mole of $N_{2}O_{4}(g)$.

Step3: Calculate the sum of $\Delta_{f}H^{\circ}$ for products

$\sum n_{products}\Delta_{f}H^{\circ}_{products}=2\times\Delta_{f}H^{\circ}(NO_{2}(g)) + 2\times\Delta_{f}H^{\circ}(H_{2}O(l))=2\times33.2+2\times(- 285.8)=66.4-571.6=-505.2$ kJ/mol

Step4: Calculate the sum of $\Delta_{f}H^{\circ}$ for reactants

$\sum n_{reactants}\Delta_{f}H^{\circ}_{reactants}=\Delta_{f}H^{\circ}(N_{2}H_{4}(l))+\Delta_{f}H^{\circ}(N_{2}O_{4}(g))=50.6 + 9.16=59.76$ kJ/mol

Step5: Calculate $\Delta_{r}H^{\circ}$

$\Delta_{r}H^{\circ}=-505.2-59.76=-564.96\approx - 565.0$ kJ/mol

Answer:

$-565.0$ kJ/mol