Question

1. hydrazine (n₂h₄) is a fuel used by some spacecraft. it is normally oxidized by n₂o₄ according to the equation: n₂h₄(l) + n₂o₄(g) → 2 n₂o(g) + 2 h₂o(g) calculate δᵣh° for this reaction using standard enthalpies of formation.

Explanation:

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

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

Step2: Identify the standard - enthalpies of formation values

The standard - enthalpy of formation values ($\Delta_{f}H^{\circ}$) are:

• $\Delta_{f}H^{\circ}(N_{2}H_{4},l) = 50.6\ kJ/mol$
• $\Delta_{f}H^{\circ}(N_{2}O_{4},g)=9.16\ kJ/mol$
• $\Delta_{f}H^{\circ}(N_{2}O,g)=82.05\ kJ/mol$
• $\Delta_{f}H^{\circ}(H_{2}O,g)= - 241.82\ kJ/mol$

Step3: Calculate the sum of the enthalpies of formation of products

For the products:
$n_{N_{2}O} = 2$, $n_{H_{2}O}=2$
$\sum n_{p}\Delta_{f}H^{\circ}(products)=2\times\Delta_{f}H^{\circ}(N_{2}O,g)+2\times\Delta_{f}H^{\circ}(H_{2}O,g)$
$=2\times82.05 + 2\times(-241.82)$
$=164.1-483.64=-319.54\ kJ/mol$

Step4: Calculate the sum of the enthalpies of formation of reactants

For the reactants:
$n_{N_{2}H_{4}} = 1$, $n_{N_{2}O_{4}} = 1$
$\sum n_{r}\Delta_{f}H^{\circ}(reactants)=\Delta_{f}H^{\circ}(N_{2}H_{4},l)+\Delta_{f}H^{\circ}(N_{2}O_{4},g)$
$=50.6 + 9.16=59.76\ kJ/mol$

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

$\Delta_{r}H^{\circ}=\sum n_{p}\Delta_{f}H^{\circ}(products)-\sum n_{r}\Delta_{f}H^{\circ}(reactants)$
$=-319.54-59.76=-379.3\ kJ/mol$

Answer:

$-379.3\ kJ/mol$