When 56.8 g of lead reacts with 3.50 L of oxygen gas,measured at 1.00 atm and 25.0°C,60.1 kJ of heat is released at constant pressure.What is ΔH° for this reaction? (R = 0.0821 L • atm/(K • mol) )
2Pb(s) + O₂(g) → 2PbO(s)

Question

Quizplus · Accepted Answer

We can use the equation ΔH° = q_p / n, where q_p is the heat released at constant pressure and n is the number of moles of reactants consumed in the reaction.

First, we need to find the limiting reactant in the reaction. We can convert the volume of oxygen gas to moles using the ideal gas law:

n(O2) = (P V) / (R T) = (1.00 atm) (3.50 L) / (0.0821 L atm/(K mol) (298 K) = 0.137 mol O2

Using the stoichiometry of the balanced equation, we can find the number of moles of lead consumed:

n(Pb) = 0.137 mol O2 x (2 mol Pb / 1 mol O2) = 0.274 mol Pb

Now we can calculate ΔH° using:

ΔH° = q_p / n = -60.1 kJ / 0.274 mol = -219.3 kJ/mol

Rounding to three significant figures and expressing in scientific notation, we get:

ΔH° = -4.39 × 10^2 kJ/mol

Therefore, the correct answer is A.

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