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?

Question

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?

Wyatt Atkins · Accepted Answer

It probably has something to do with the limiting reagent you chose and the enthalpy of reaction value you came up with.

How much product you make is determined by the limiting reagent, which is the one that is used up first. Here is a fairly quick method to identify which reagent it is.

THE REAGENT LIMITATION

You see how the equation you gave asks for one equivalent of #"B"_2"H"_6(g)# (diborane)? Now compare to how many equivalents you need of #"Cl"_2(g)# relative to the amount of #"B"_2"H"_6(g)#.

The coefficient in front of #"B"_2"H"_6# implicitly says #1#, but the one in front of #"Cl"_2(g)# says #6#... you need six equivalents.

Now, do you ACTUALLY have six times the number of #"mol"#s #"Cl"_2(g)# as you have #"B"_2"H"_6(g)#? No, you have #0.252/0.111 = \mathbf(2.27)# times as much as you have #"B"_2"H"_6#.

The limiting reagent must therefore be chlorine because there isn't enough of it to react with that amount of diborane.

MOLAR ENTHALPY OF REACTION #\mathbf(DeltaH_"rxn"^@)# VS HEAT FLOW #\mathbf(q)#

Ultimately, we need to determine the heat #\mathbf(q)# that is released from this reaction.

We have the molar enthalpy of reaction defined for #25^@ "C"# and #"1 bar"# (not #0^@ "C"#!!), #color(green)(DeltaH_"rxn"^@ = -"1396 kJ/mol B"_2"H"_6(g))#.

It is defined this way because the reaction you are referencing is based on #\mathbf("1 mol")# of #\mathbf("B"_2"H"_6)# and conventionally, the proper units are #"kJ/mol"#.

The units are NOT just #"kJ"#. That will matter!

The heat energy released during a reaction under constant pressure divided by the number of moles of the limiting reagent is the molar enthalpy of reaction.

That's where the limiting reagent enters the picture. :)

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

It is suggested that you can:

Conveniently use a referenced #DeltaH_"rxn"^@# from thermodynamic tables, which is on a per-mol basis (hence, #"kJ/mol"#), defined for specific atmospheric conditions (#25^@ "C"# and #"1 bar"#), and based on the particular reagent which has a stoichiometric coefficient of #\mathbf(1)#. In this case it is #"B"_2"H"_6(g)#.

Recognize that the enthalpy as-written (#-"1396 kJ/mol"#) is still in reference to #\mathbf("B"_2"H"_6)# being the limiting reagent, but we know that it is NOT the limiting reagent. So, we convert it to accommodate for that.

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 #"1 mol"# of #"Cl"_2(g)#, but #"0.252 mol"#.

DROPPING THE NORMSIDEAL REACTION TO YOURS

Since we currently are not talking about #"B"_2"H"_6# as the limiting reagent, but about #"Cl"_2(g)#, we need to convert to get the accurate value for the molar enthalpy of reaction that is relative to #"Cl"_2(g)#.

#color(green)(DeltaH_"rxn"^@) = (-"1396 kJ")/(cancel("1 mol B"_2"H"_6(g))) xx (cancel("1 mol B"_2"H"_6(g)))/("6 mols Cl"_2(g))#

#= color(green)(-"232.67 kJ/mol")#

(of #color(green)("Cl"_2(g))#)

That was the first step to the scaling: making sure you are referring to #\mathbf("1 mol")# of limiting reagent in your standard reaction.

The enthalpy of reaction can then be scaled down to the molar heat released.

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

#q_"rxn" = DeltaH_"rxn"^@ xx "mols Limiting Reagent"#

#color(blue)(q_"rxn") = -"232.67 kJ/"cancel("mol Cl"_2(g)) xx "0.252" cancel("mols Cl"_2(g))#

For this second scaling, you should see that we basically multiplied the enthalpy in #"kJ"# by #0.252/1#, and you had #"0.252 mol"#s of #"Cl"_2(g)#.

Consequently, we have reduced the response twice:

The reaction's heat is as follows:

#color(blue)(q_"rxn") =# #color(blue)(-"58.63 kJ")#

Or, you could say that #\mathbf(58.63 kJ)# of heat was released.

Caleb Adams · Answer

undefined The heat of reaction ($ \Delta H_{\text{rxn}} $) can be calculated using the given moles of reactants and the given value for $ \Delta H_{\text{rxn}} $.

First, determine the limiting reactant, which is $ B_2H_6 $.

Then, use the stoichiometry of the reaction to calculate the moles of $ H_2 $ produced.

Finally, apply the formula: $ \Delta H_{\text{rxn}} = \text{moles of } H_2 \times \Delta H_{\text{rxn}} $ to find the heat of reaction.

For the reaction #"B"_2"H"_6(g) + 6"Cl"_2(g) -&gt; "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?

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?