# Question 56be2

Mar 10, 2016

The reaction will give off $\text{100 kJ}$ of heat.

#### Explanation:

The problem provides you with the thermochemical equation for the decomposition of hydrogen peroxide, ${\text{H"_2"O}}_{2}$

$\textcolor{red}{2} \text{H"_2"O"_text(2(aq]) -> 2"H"_2"O"_text((l]) + "O"_text(2(g])" ", DeltaH_text(rxn) = -"200 kJ}$

Now, a thermochemical equation tells you the change in enthalpy for the reaction consumes the number of moles of reactant(s) and produces the number of moles of products that appear listed as stoichiometric coefficients in the balanced chemical equation.

The thermochemical equation given to you can be interpreted like this:

When $\textcolor{red}{2}$ moles of hydrogen peroxide decompose to form $2$ moles of water and $1$ mole of oxygen gas, the change in enthalpy for the reaction is equal to $\Delta {H}_{\text{rxn" = -"200 kJ}}$

Since you're interested in looking at this from the perspective of hydrogen peroxide, your reactant, you can say that when $\textcolor{red}{2}$ moles of hydrogen peroxide decompose, $\text{200 kJ}$ of heat are being given off by the reaction.

Remember, the minus sign in $\Delta {H}_{\text{rxn}}$ is used to denote heat given off.

So, you know how much heat is being given off when $\textcolor{red}{2}$ moles of hydrogen peroxide decompose, but how much heat will be released when $\text{34 g}$ of hydrogen peroxide decompose?

Use the compound's molar mass to calculate how many moles you have in that $\text{34-g}$ sample

34color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O"_2)/(34.015color(red)(cancel(color(black)("g")))) ~~ "1.0 moles H"_2"O"_2#

So, if $\text{200 kJ}$ are given off when two moles of hydrogen peroxide decompose, it follows that the decomposition of one mole of hydrogen peroxide will give off

$1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mole H"_2"O"_2))) * overbrace("200 kJ"/(color(red)(2)color(red)(cancel(color(black)("moles H"_2"O"_2)))))^(color(purple)("given by thermochemical equation")) = color(green)(|bar(ul(color(white)(a/a)"100 kJ} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, when $\text{34 g}$ of hydrogen peroxide decompose, $\text{100 kJ}$ are being given off by the reaction. This is equivalent to saying that when one mole of hydrogen peroxide undergoes decomposition, the enthalpy change of reaction will be

$\Delta {H}_{\text{rxn 1 mole" = -"100 kJ}}$