Consider the reaction: 2al(s) + fe2o3(s) → al2o3(s) + 2fe(s) the δhf for fe2o3(s) = -824.3 kj/mole. the δhf for al2o3(s) = -1675.7 kj/mole. finish the equation. δhrxn = [(1)( kj/mole) + (2)( kj/mole)] - [(1)( kj/mole) + (2) ( kj/mole)]

for the given reaction is -851.4 kJ/mol.

Explanation:

Enthalpy of the reaction is equal to the total sum of the standard enthalpies of the  formation of products minus the total sum of the standard enthalpies of formation of reactants. It is represented by

Mathematically,

For the given reaction:

The standard enthalpy of the elements present in their standard state is always zero.

for the reaction is calculated by:

We are given:

Putting values in above equation, we get:

Hence, for the given reaction is -851.4 kJ/mol.

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

just did the assignment and the answer was #1 +2 n