Question

Manganese metal can be obtained by the reaction of manganese dioxide with aluminium. 4Al(s)+3MnO2(s) 2 Al2O3(s)+3 Mn(s) Calculate the enthalpy change of reaction. Given: 2Al(s)+3/2 O2(8) AO3(s) AH-1680 kJ mol Mn(s)+O2(g) MnO2(s) -520 kJ mol1. AH

Answer 1

Answer: $\Delta H=-1800\frac{kJ}{mol}$

Explanation:

We have the following reaction:

$4Al(s)+3MnO_2(s) \rightarrow Al_2O_3(s)+3 Mn(s)$

And we are given thermodinamic data from the two following reactions:

$2Al(s)+\frac{3}{2}  O_2(8) \rightarrow Al_2O_3(s)\ \Delta H=-1680 \frac{kJ}{mol}\\Mn(s)+O_2(g)\ \rightarrow MnO_2(s) \Delta H=-520 \frac{kJ}{mol}$

According to Hess's Law, if we can somehow combine both ecuations to get the first one, we can sum both $\Delta H$ to get the total change of enthalpy.

We identify that in the second ecuation, $MnO_2$ is on the right side, and we want it to be on the left side. We can invert that ecuation and that means getting the opposite $\Delta H$:

$MnO_2(s)\rightarrow Mn(s)+O_2(g) \ \Delta H=520 \frac{kJ}{mol}$

Now, before combining both ecuations, we must get the same stoichometric coefficents present on the one we want to obtain. For that, just multiply the first one by 2 and the second one by 3, that means also multipying the corresponding $\Delta H$:

$4Al(s)+3O_2(8) \rightarrow 2Al_2O_3(s)\ \Delta H=-3360\\3MnO_2(s)\rightarrow 3Mn(s)+3O_2(g) \ \Delta H=1560 \frac{kJ}{mol}$

Now, just sum both ecuations to get the original one:

$\ \ \  4Al(s)+\cancel{3O_2(g)} \rightarrow 2Al_2O_3(s)\ \ \ \Delta H=-3360\frac{kJ}{mol}\\+3MnO_2(s)\rightarrow 3Mn(s)+\cancel{3O_2(g)} \ \Delta H=1560 \frac{kJ}{mol}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\\ \ \ \ 4Al(s)+3MnO_2(s) \rightarrow Al_2O_3(s)+3 Mn(s)\ \ \Delta H=-1800\frac{kJ}{mol}$