Using the information below,calculate $\Delta$ H_f° for PbO(s) PbO(s) + CO(g) $\to$ Pb(s) + CO₂(g) $\Delta$ H° = -131.4 kJ
$\Delta$ H_f° for CO₂(g) = -393.5 kJ/mol
$\Delta$ H_f° for CO(g) = -110.5 kJ/mol

Question

Using the information below,calculate

\Delta

H_f° for PbO(s) PbO(s) + CO(g)

\to

Pb(s) + CO₂(g)

\Delta

H° = -131.4 kJ

\Delta

H_f° for CO₂(g) = -393.5 kJ/mol

\Delta

H_f° for CO(g) = -110.5 kJ/mol

Quizplus · Accepted Answer

To calculate the standard enthalpy of formation for Pb(s), we need to use the following equation: PbO(s) + CO(g) → Pb(s) + CO₂(g) We know: ∆fH° for PbO(s) is unknown ∆fH° for CO₂(g) = -393.5 kJ/mol (given) ∆fH° for CO(g) = -110.5 kJ/mol (given) ∆fH° for Pb(s) is the value we want to find We can use Hess's Law here, which states that the sum of the enthalpy changes for a series of reactions is equal to the enthalpy change of the overall reaction. So, we can write: ∆fH° for Pb(s) = ∆fH° for PbO(s) + (-∆fH° for CO(g)) + (-∆fH° for CO₂(g)) Plugging in the values we know: -∆fH° for CO(g) = -(-110.5 kJ/mol) = 110.5 kJ/mol (note the negative sign from the equation) -∆fH° for CO₂(g) = -393.5 kJ/mol ∆fH° for Pb(s) = ∆fH° for PbO(s) + 110.5 kJ/mol + 393.5 kJ/mol ∆fH° for Pb(s) = ∆fH° for PbO(s) + 504.0 kJ/mol Now, we can use the given value of ∆fH° for the overall reaction, which is -131.4 kJ/mol, and plug in the enthalpies we just calculated: -131.4 kJ/mol = ∆fH° for PbO(s) + 504.0 kJ/mol ∆fH° for PbO(s) = -(131.4 kJ/mol + 504.0 kJ/mol) = -635.4 kJ/mol Therefore, the value we want is the opposite of the value we just calculated, because we want the enthalpy of formation for PbO(s) and not for Pb(s): ∆fH° for PbO(s) = 635.4 kJ/mol And finally, we can calculate the standard enthalpy change of the reaction: ∆H° = ∆H°f(Pb(s)) + ∆H°f(CO₂(g)) - ∆H°f(PbO(s)) - ∆H°f(CO(g)) ∆H° = 0 + (-393.5 kJ/mol) - (-635.4 kJ/mol) - (-110.5 kJ/mol) ∆H° = -151.6 kJ/mol Therefore, the answer is A) -151.6 kJ/mol.

Using the Information Below,calculate Δ\DeltaΔ Hf° for PbO(s) PbO(s)+ CO(g) →\to→ Pb(s)+ CO2(g)

Using the Information Below,calculate $\Delta$ H_f° for PbO(s) PbO(s)+ CO(g) $\to$ Pb(s)+ CO₂(g)