When #"32.4 mL"# of liquid benzene (#C_6H_6#, #rho = "0.879 g/mL"#) reacts with #"81.6 L"# of oxygen gas, measured at #"1.00 atm"# and #"25°C"#, #1.19xx10^3 "kJ"# of heat is released at const. pressure. What is #DeltaH^@# for the following reaction?

To find ΔH° for the reaction, we need to calculate the moles of benzene and oxygen reacting and then use the given heat released to determine the enthalpy change.

First, we need to convert mL of benzene to grams using its density:

32.4 mL * 0.879 g/mL = 28.4776 g

Next, we convert grams of benzene to moles using its molar mass:

28.4776 g / 78.11 g/mol = 0.3645 mol

Now, we calculate moles of oxygen using the ideal gas law:

PV = nRT

(1.00 atm) * (81.6 L) = n * (0.0821 L atm/mol K) * (298 K)

n = (1.00 atm * 81.6 L) / (0.0821 L atm/mol K * 298 K)
n ≈ 3.448 mol

Since the reaction ratio is 1:15 (from the balanced chemical equation), the limiting reactant is benzene. Therefore, all the oxygen reacts with 0.3645 mol of benzene.

Now, we use the given heat released:

1.19 × 10^3 kJ = q (at constant pressure)

Finally, we can calculate ΔH° using the formula:

ΔH° = q / n

ΔH° = (1.19 × 10^3 kJ) / 0.3645 mol

ΔH° ≈ 3269.526 kJ/mol

Therefore, ΔH° for the reaction is approximately 3269.526 kJ/mol.