Manganese dioxide reacts with aluminum to form aluminum oxide and manganese according to the equation below.

The answer is -1839.5 kJ/mol. This was calculated using the given formula for ΔH. You have to calculate the sum of ΔHf for products and subtract from it the ΔHf of the reactants.

ΔHrxn = (2 mol *-1699.8 kJ/mol) - (3 mol * -520 kJ/mol) where 2 mol Al2O3 and 3 mol MnO2 came from the given reaction. The other reactant and products are in their standard form so their ΔHf is zero.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm