Question

Manganese metal can be obtained by the reaction of manganese dioxide with aluminium. 4Al(s) + 3MnOz(s) → 2 Al2O3(s) + 3 Mn(s) Calculate the enthalpy change of reaction. Given: 2Al(s) + 3/2 O2(g) →Al2O3(s) AH = -1680 kJ mol'. Mn(s) + O2(g) → MnO2(s) AH = -520 kJ mol!.

Answer 1

Answer : The enthalpy change of reaction is -1800 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given final reaction is,

$4Al(s)+3MnO_2(s)\rightarrow 2Al_2O_3(s)+3Mn(s)$    $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $2Al(s)+\frac{3}{2}O_2(g)\rightarrow Al_2O_3(s)$    $\Delta H_1=-1680kJ/mole$

(2) $Mn(s)+O_2(g)\rightarrow MnO_2(s)$    $\Delta H_2=-520kJ/mole$

First we will multiply reaction 1  by 2 and reverse reaction of reaction 2 by 3 then adding both the equation, we get :

The expression for final enthalpy is,

$\Delta H=[n\times \Delta H_1]+[n\times (-\Delta H_2)]$

where,

n = number of moles

$\Delta H=[2mole\times (-1680kJ/mole)]+[3\times -(-520kJ/mole)]$

$\Delta H=-1800kJ$

Therefore, the enthalpy change of reaction is -1800 kJ