Question

Al2(SO4)3(s) + H2O(l) ---- Al2O3(s) + H2SO4 (aq) calculate enthalpy formation for this reaction. Balance the reaction. Calculate the total enthalpy change that would occur from 15 moles of Al2(SO4)3

Answer 1

The given chemical equation is:

`Al_(2)(SO_(4))_(3)(aq)+H_(2)O(l)-->Al_(2)O_(3)(aq)+H_(2)SO_(4)(aq)`

On balancing the equation we get,

`Al_(2)(SO_(4))_(3)(aq)+3H_(2)O(l)-->Al_(2)O_(3)(aq)+3H_(2)SO_(4)(aq)`

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:

Δ
`H_(reaction)^(0)</p><p>=[H_(f)^(0)(Al_(2)O_(3)(s)) + (3*H_(f)^(0)(H_(2)SO_(4)(aq))] -   [H_(f)^(0)(Al_(2)SO_(4)(aq)) + (3*H_(f)^(0)(H_(2)O(l))]`

=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]

=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]

=-98.21kJ/mol

Total enthalpy change when 15 mol of
`Al_(2)(SO_(4))_(3)`reacts will be=

`15 mol Al_(2)(SO_(4))_(3)*(-98.21kJ)/(1 molAl_(2)(SO_(4))_(3)) =-1473.15kJ/mol`