Question

I have tried initiating the question but I am not sure how to continue or what to do after moving things to the other side

Answer 1

Answer:

+828kJ

Explanations:

Given the following chemical reaction with their heat of reactions

$\begin{gathered} 2Al(s)+\frac{3}{2}O_2(g)\rightarrow Al_2O_3(s);\text{ }\triangle H=-1675.7kJ..............\text{ 1} \\ CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g);\text{ }\triangle H=-282.7kJ...................2 \end{gathered}$

In order to arrive at the resulting reaction, we will carry out the followiing transformation;

• Swap the reactant with product, of equation 1 and;

,

• Multiply equation 2 by 3, to have;

$\begin{gathered} Al_2O_3(s)\rightarrow2Al(s)+\frac{3}{2}O_2(g);\text{ }\triangle H=+1675.7kJ \\ 3CO(g)+\frac{3}{2}O_2(g)\rightarrow3CO_2(g);\text{ }\triangle H=-3(282.7)=-848.1kJ \end{gathered}$

Cancel out the Oxygen element in both equation and add to have;

$Al_2O_3(s)+3CO(g)\rightarrow2Al(s)+3CO_2(g);\text{ }\triangle H_{rxn}=+1675.7-848.1=+827.6kJ$

Hence the enthalpy change of the reaction to 3sf is +828kJ