# For the reaction "B"_2"H"_6(g) + 6"Cl"_2(g) -> "B"_2"Cl"_6(g) + 6"H"_2(g), if "0.252 mols Cl"_2 reacts with "0.111 mols B"_2"H"_6, and DeltaH_(rxn)^@ = -"1396 kJ", what is the heat of reaction?

Jan 28, 2016

I'm going to guess it has something to do with which reagent you used as the limiting reagent, and the value you used for your enthalpy of reaction.

The limiting reagent, the one that is used up first, dictates how much product you will make. Here is a fairly quick way to figure out which reagent it is.

THE LIMITING REAGENT

You see how the equation you gave asks for one equivalent of ${\text{B"_2"H}}_{6} \left(g\right)$ (diborane)? Now compare to how many equivalents you need of ${\text{Cl}}_{2} \left(g\right)$ relative to the amount of ${\text{B"_2"H}}_{6} \left(g\right)$.

The coefficient in front of ${\text{B"_2"H}}_{6}$ implicitly says $1$, but the one in front of ${\text{Cl}}_{2} \left(g\right)$ says $6$... you need six equivalents.

Now, do you ACTUALLY have six times the number of $\text{mol}$s ${\text{Cl}}_{2} \left(g\right)$ as you have ${\text{B"_2"H}}_{6} \left(g\right)$? No, you have $\frac{0.252}{0.111} = \setminus m a t h b f \left(2.27\right)$ times as much as you have ${\text{B"_2"H}}_{6}$.

So, chlorine must be the limiting reagent. You don't have enough of it to react with that much diborane.

MOLAR ENTHALPY OF REACTION $\setminus m a t h b f \left(\Delta {H}_{\text{rxn}}^{\circ}\right)$ VS HEAT FLOW $\setminus m a t h b f \left(q\right)$

Ultimately, we need to determine the heat $\setminus m a t h b f \left(q\right)$ that is released from this reaction.

We have the molar enthalpy of reaction defined for ${25}^{\circ} \text{C}$ and $\text{1 bar}$ (not ${0}^{\circ} \text{C}$!!), $\textcolor{g r e e n}{\Delta {H}_{\text{rxn"^@ = -"1396 kJ/mol B"_2"H}} _ 6 \left(g\right)}$.

It is defined this way because the reaction you are referencing is based on $\setminus m a t h b f \left(\text{1 mol}\right)$ of $\setminus m a t h b f \left({\text{B"_2"H}}_{6}\right)$ and conventionally, the proper units are $\text{kJ/mol}$.

The units are NOT just $\text{kJ}$. That will matter!

The molar enthalpy of reaction is equal to the heat energy released from the reaction in constant pressure conditions divided by the $m o l$s of limiting reagent.

And there we go, that's where the limiting reagent comes in. :)

\mathbf(DeltaH_"rxn"^@ = (q_"rxn")/("mols Limiting Reagent"))

The idea is that you can:

1. Conveniently use a referenced $\Delta {H}_{\text{rxn}}^{\circ}$ from thermodynamic tables, which is on a per-mol basis (hence, $\text{kJ/mol}$), defined for specific atmospheric conditions (${25}^{\circ} \text{C}$ and $\text{1 bar}$), and based on the particular reagent which has a stoichiometric coefficient of $\setminus m a t h b f \left(1\right)$. In this case it is ${\text{B"_2"H}}_{6} \left(g\right)$.

2. Recognize that the enthalpy as-written ($- \text{1396 kJ/mol}$) is still in reference to $\setminus m a t h b f \left({\text{B"_2"H}}_{6}\right)$ being the limiting reagent, but we know that it is NOT the limiting reagent. So, we convert it to accommodate for that.

3. Scale it down to the size of your real reaction at the same atmospheric conditions to accommodate for the fact that you are not using exactly $\text{1 mol}$ of ${\text{Cl}}_{2} \left(g\right)$, but $\text{0.252 mol}$.

SCALING THE STANDARD REACTION DOWN TO YOUR REACTION

Since we currently are not talking about ${\text{B"_2"H}}_{6}$ as the limiting reagent, but about ${\text{Cl}}_{2} \left(g\right)$, we need to convert to get the accurate value for the molar enthalpy of reaction that is relative to ${\text{Cl}}_{2} \left(g\right)$.

$\textcolor{g r e e n}{\Delta {H}_{\text{rxn"^@) = (-"1396 kJ")/(cancel("1 mol B"_2"H"_6(g))) xx (cancel("1 mol B"_2"H"_6(g)))/("6 mols Cl}} _ 2 \left(g\right)}$

$= \textcolor{g r e e n}{- \text{232.67 kJ/mol}}$

(of $\textcolor{g r e e n}{{\text{Cl}}_{2} \left(g\right)}$)

That was the first step to the scaling: making sure you are referring to $\setminus m a t h b f \left(\text{1 mol}\right)$ of limiting reagent in your standard reaction.

Next, we can scale the enthalpy of reaction down to the molar heat released.

(q_"rxn")/"mols Limiting Reagent" = DeltaH_"rxn"^@

${q}_{\text{rxn" = DeltaH_"rxn"^@ xx "mols Limiting Reagent}}$

$\textcolor{b l u e}{{q}_{\text{rxn") = -"232.67 kJ/"cancel("mol Cl"_2(g)) xx "0.252" cancel("mols Cl}} _ 2 \left(g\right)}$

For this second scaling, you should see that we basically multiplied the enthalpy in $\text{kJ}$ by $\frac{0.252}{1}$, and you had $\text{0.252 mol}$s of ${\text{Cl}}_{2} \left(g\right)$.

So, what we've done is scaled the reaction down twice:

1. So that the amount of chlorine, the limiting reagent, referred to by the enthalpy of reaction is $\text{1 mol}$ instead of $\text{6 mol}$s.
2. So that the amount of chlorine used is $\text{0.252 mol}$ instead of $\text{1 mol}$.

The heat of reaction is thus:

$\textcolor{b l u e}{{q}_{\text{rxn}}} =$ $\textcolor{b l u e}{- \text{58.63 kJ}}$

Or, you could say that $\setminus m a t h b f \left(58.63 k J\right)$ of heat was released.