Question

manganese dioxide mno₂(s), δh_f = -520.0 kj reacts with aluminum to form aluminum oxide al₂o₃(s), δh_f = -1699.8 kj and manganese according to the equation.
3mno₂(s) + 4al(s) → 2al₂o₃(s) + 3mn(s)
what is the enthalpy of the reaction?
use δh_rex = σ(δh_f, products) - σ(δh_f, reactants).
1,179.8 kj -1,839.6 kj -1,179.8 kj
1,839.6 kj

Explanation:

Step1: Identify products and reactants

Products: \(2\ce{Al2O3}(s)\), \(3\ce{Mn}(s)\)
Reactants: \(3\ce{MnO2}(s)\), \(4\ce{Al}(s)\)

Step2: Recall \(\Delta H_f\) for elements

\(\Delta H_f\) for pure elements (\(\ce{Al}(s)\), \(\ce{Mn}(s)\)) is \(0\ \text{kJ}\).

Step3: Calculate \(\sum \Delta H_{f, \text{products}}\)

For \(2\ce{Al2O3}(s)\): \(2 \times (-1699.8\ \text{kJ}) = -3399.6\ \text{kJ}\)
For \(3\ce{Mn}(s)\): \(3 \times 0 = 0\ \text{kJ}\)
Sum: \(-3399.6 + 0 = -3399.6\ \text{kJ}\)

Step4: Calculate \(\sum \Delta H_{f, \text{reactants}}\)

For \(3\ce{MnO2}(s)\): \(3 \times (-520.0\ \text{kJ}) = -1560.0\ \text{kJ}\)
For \(4\ce{Al}(s)\): \(4 \times 0 = 0\ \text{kJ}\)
Sum: \(-1560.0 + 0 = -1560.0\ \text{kJ}\)

Step5: Apply \(\Delta H_{\text{rxn}}\) formula

\(\Delta H_{\text{rxn}} = \sum \Delta H_{f, \text{products}} - \sum \Delta H_{f, \text{reactants}}\)
\(\Delta H_{\text{rxn}} = (-3399.6) - (-1560.0) = -3399.6 + 1560.0 = -1839.6\ \text{kJ}\)

Answer:

\(-1,839.6\ \text{kJ}\) (corresponding to the option with this value)