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3 years ago

Find ΔHrxn for the following reaction: 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)

Chemistry

1 answer:

ch4aika [34]3 years ago

5 0

Answer:

ΔH°rxn = -827.5 kJ

Explanation:

Let's consider the following balanced equation.

2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)

We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]

ΔH°rxn = -827.5 kJ

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$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

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