Question

Consider the following intermediate chemical equations.

Answer 1

Answer:

        $Overall\text{ }enthalpy=-1,300kJ$

Explanation:

According to Hess’s law, you can calculate the enthalpy  of the overall chemical reaction by adding enthalpy changes of the intermediate reactions, at the same temperature.

First, you must find how the intermediate equations form the overall equation.

Here, the overall equation is:

            $P_4O_{6}(s)+2O_2(g)\rightarrow P_4O_{10}(s)$

To obtain it you can swift both the first and second intermediate equations and sum them. When you swift a chemcial equation, the corresponding enthalpy change has the opposite sign:

Change

              $P_4(s)+3O_2(g)\rightarrow P_4O_6(s)$              $\Delta H_1=-1,640kJ$

     to

         $P_4O_6(s)\rightarrow P_4(s)+3O_2(g)$            $-\Delta H_1=1,640kJ$

Change

             $P_4O_{10}(s)\rightarrow P_4(s)+5O_2(g)$          $\Delta H_2=2,940kJ$

    to

             $P_4(s)+5O_2(g)\rightarrow P_4O_{10}(s)$         $-\Delta H_2=-2,940kJ$

Now add the two transformed equations:

             $P_4O_{6}(s)+2O_2(g)\rightarrow P_4O_{10}(s)$

$Overall\text{ }enthalpy=-\Delta H_1+-\Delta H_2=1,649kJ+(-2,940.1kJ)=-1300.1kJ$

$Overall\text{ }enthalpy=-1,300kJ$