Some spacecrafts use hydrazine (N2H4) as fuel and is burned according to the following reaction:

N2H4(l)+N2O4(g)=>2 N2O(g)+2 H2O(g)

Calculate the DHrxn° for this reaction using the standard enthalpies
of formation from Appendix IIB. Calculate your answer to 4 sig figs

dHrxn = (n*dHf products) - (n*Hf reactants)

To calculate the standard enthalpy change (ΔHrxn°) for the given reaction, we will use the standard enthalpies of formation (ΔHf°) from Appendix IIB. The reaction is as follows:

N2H4(l) + N2O4(g) → 2 N2O(g) + 2 H2O(g)

We need to find the ΔHf° values for each compound involved in the reaction. The values are:

ΔHf° (N2H4) = 95.6 kJ/mol
ΔHf° (N2O4) = 9.16 kJ/mol
ΔHf° (N2O) = 81.6 kJ/mol
ΔHf° (H2O) = -285.83 kJ/mol

Now we can use the following equation to calculate the ΔHrxn°:

ΔHrxn° = ΣnΔHf° (products) - ΣmΔHf° (reactants)

Let's calculate ΔHrxn°:

ΔHrxn° = (2 mol × ΔHf° (N2O)) + (2 mol × ΔHf° (H2O)) - (ΔHf° (N2H4)) - (ΔHf° (N2O4))
ΔHrxn° = (2 mol × 81.6 kJ/mol) + (2 mol × -285.83 kJ/mol) - 95.6 kJ/mol - 9.16 kJ/mol
ΔHrxn° = 163.2 kJ - 571.66 kJ - 95.6 kJ - 9.16 kJ
ΔHrxn° = -513.42 kJ

Therefore, the ΔHrxn° for the given reaction is -513.42 kJ.

To calculate the standard enthalpy change (ΔHrxn°) for the given reaction, we need to calculate the overall change in enthalpy (∆H°) based on the standard enthalpies of formation (∆Hf°) for each compound involved.

Step 1: Find the standard enthalpies of formation (∆Hf°) for each compound involved in the reaction from Appendix IIB. The standard enthalpies of formation for N2O(g), H2O(g), N2H4(l), and N2O4(g) are as follows:

∆Hf°(N2O(g)) = 82.05 kJ/mol
∆Hf°(H2O(g)) = -241.82 kJ/mol
∆Hf°(N2H4(l)) = 50.60 kJ/mol
∆Hf°(N2O4(g)) = 9.16 kJ/mol

Step 2: Determine the stoichiometric coefficients in the balanced equation:

N2H4(l) + N2O4(g) --> 2 N2O(g) + 2 H2O(g)

According to the balanced equation, the stoichiometric coefficients for N2H4(l) and N2O4(g) are both 1, while the coefficients for N2O(g) and H2O(g) are both 2.

Step 3: Calculate the overall change in enthalpy (∆H°) by summing the products and subtracting the reactants:

∆H° = (2 * ∆Hf°(N2O(g))) + (2 * ∆Hf°(H2O(g))) - (∆Hf°(N2H4(l))) - (∆Hf°(N2O4(g)))

∆H° = (2 * 82.05 kJ/mol) + (2 * -241.82 kJ/mol) - (50.60 kJ/mol) - (9.16 kJ/mol)

Step 4: Calculate the value of ∆H° using the given standard enthalpies of formation:

∆H° = 164.10 kJ/mol - 483.64 kJ/mol - 50.60 kJ/mol - 9.16 kJ/mol

∆H° = -379.30 kJ/mol

Therefore, the ΔHrxn° for this reaction is -379.30 kJ/mol.