# For the reaction: 2H_2O + 137 kcal -> 2H_2(g) + O_2(g) how many kcal are needed to form 2.00 moles O_2(g)?

Dec 11, 2016

$\text{274 kcal}$

#### Explanation:

Start by taking a look at the thermochemical equation given to you

$2 {\text{H"_ 2"O"_ ((l)) + "137 kcal" -> 2"H"_ (2(g)) + "O}}_{2 \left(g\right)}$

Notice that this thermochemical equation describes an endothermic reaction. You can say that because you have a heat term added to the reactants' side.

This essentially tells you that heat is needed in order for the reaction to take place. This can also be shown by using the enthalpy change of reaction, $\Delta {H}_{\text{rxn}}$, which in this case is equal to

$\Delta {H}_{\text{rxn" = +"137 kcal}}$

The enthalpy change of reaction is positive because the reaction is endothermic

Now, take a look at the stoichiometric coefficients. You can say that it takes $\text{137 kcal}$ of heat in order to get $2$ moles of water to produce $2$ moles of hydrogen gas and $1$ mole of oxygen gas.

The problem wants you to figure out how much heat is needed in order for the reaction to produce $2.00$ moles of oxygen gas.

You can use the fact that $\text{137 kcal}$ of heat are needed in order to produce $1$ mole of oxygen gas to say that

$2.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles O"_2))) * overbrace("137 kcal"/(1color(red)(cancel(color(black)("mole O"_2)))))^(color(blue)("given by the thermochemical equation")) = color(darkgreen)(ul(color(black)("274 kcal}}}}$

The answer is rounded to three sig figs.