A The standard molar enthalpy of formation of diborane, B2 H6(g), cannot be determined directly

A The standard molar enthalpy of formation of diborane, B2...

A The standard molar enthalpy of formation of diborane, $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g}),$ cannot be determined directly because the compound cannot be prepared by the reaction of boron and hydrogen. It can be calculated from other enthalpy changes, however. The following enthalpy changes can be measured. $4 \mathrm{B}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})$ $\Delta_{1} H^{\circ}=-2543.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{pxn}$ $\mathrm{H}_{2}(\mathrm{g})+^{1 / 2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ $\Delta_{r} H^{\prime \prime}=-241.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}$ $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ $\Delta_{\tau} H^{\circ}=-2032.9 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}$ (a) Show how these equations can be added together to give the equation for the formation of $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})$ from $\mathrm{B}(\mathrm{s})$ and $\mathrm{H}_{2}(\mathrm{g})$ in their standard states. Assign enthalpy changes to each reaction. (b) Calculate $\Delta_{f} H^{\circ}$ for $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})$ (c) Draw an energy level diagram that shows how the various enthalpies in this problem are related. (d) Is the formation of $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})$ from its elements exo-or endothermic?

A The standard molar enthalpy of formation of diborane, $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g}),$ cannot be determined directly because the compound cannot be prepared by the reaction of boron and hydrogen. It can be calculated from other enthalpy changes, however. The following enthalpy changes can be measured.
$4 \mathrm{B}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})$
$\Delta_{1} H^{\circ}=-2543.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{pxn}$
$\mathrm{H}_{2}(\mathrm{g})+^{1 / 2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
$\Delta_{r} H^{\prime \prime}=-241.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}$
$\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
$\Delta_{\tau} H^{\circ}=-2032.9 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}$
(a) Show how these equations can be added together to give the equation for the formation of $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})$ from $\mathrm{B}(\mathrm{s})$ and $\mathrm{H}_{2}(\mathrm{g})$ in their
standard states. Assign enthalpy changes to each reaction.
(b) Calculate $\Delta_{f} H^{\circ}$ for $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})$
(c) Draw an energy level diagram that shows how the various enthalpies in this problem are related.
(d) Is the formation of $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})$ from its elements exo-or endothermic?

Key Concepts

Hess's Law

Hess's Law is the principle stating that the total enthalpy change for a chemical reaction is the same regardless of the pathway the reaction takes. This allows the calculation of formation enthalpies for compounds that cannot be synthesized directly by combining their elements, by using a series of reactions with known enthalpy changes to obtain the desired overall reaction.

Standard Enthalpy of Formation

The standard enthalpy of formation is the enthalpy change that occurs when one mole of a compound is formed from its elements in their standard states. It is a fundamental thermodynamic quantity and serves as a reference point for calculating other enthalpy changes in chemical reactions.

Reaction Enthalpy

Reaction enthalpy is the heat absorbed or released during a chemical reaction under constant pressure. It is calculated by summing the enthalpy changes of individual reactions that, when combined, yield the overall reaction. This concept is central to thermochemistry and energy calculations in chemical reactions.

Energy Level Diagram

An energy level diagram is a graphical representation that shows the relative energies of reactants and products as well as the energy changes associated with a reaction. It is used to visualize the exothermic or endothermic nature of reactions and the steps involved when applying Hess's Law.

Exothermic and Endothermic Reactions

Exothermic reactions release energy, usually in the form of heat, resulting in a negative enthalpy change, whereas endothermic reactions absorb energy, leading to a positive enthalpy change. Understanding whether a reaction is exothermic or endothermic is crucial for predicting reaction spontaneity and safety considerations in chemical processes.

Transcript

00:01 So this question is asking us to use hess's law to show that the three equations can be added together and then to equal a target reaction and then to figure out the delta -h of the reaction.

00:11 That's part a and b, which i'm going to do together at the same time, then to show them an energy diagram, which i'm going to do then next, and then to talk about the if the reaction is endorexothermic.

00:23 So hesse's law talks about just the idea that a reaction is a state function.

00:27 A state function just talks about that the pathway we get to a target reaction doesn't matter that we can simply get out the energy because the energy chain is going to be the same no matter what.

00:40 So, and whatever we do to the reactions, as we manipulate them, we have to do their delta h's because the delta h is a molar quantity.

00:48 So if we half the number of moles, we have to half the energy.

00:51 So as we begin to manipulate h1, we have to work on them with their molar quantities.

00:57 So as i work on them, i like to identify where they are.

01:01 So boron, boron, h2, h2, b2h6.

01:10 I only focus on them.

01:12 Everything else tends to work out as we work through.

01:16 So the first one, boron itself is a reaction to the coefficient of 2.

01:23 Well, it's still a reaction, but it's got a coefficient of 4.

01:26 So i have to multiply everything in this equation by one half.

01:29 So as i work through, i have to multiply everything by one half.

01:41 And that one half includes the delta h value.

01:51 Now in the very next one, we notice that h2 is a reactant in bolt, but it's got a coefficient of three.

01:58 So it's multiplied everything by three that includes the delta h value.

02:08 In the very last one, b2h6, it's got a coefficient of one in both, but in the target, it's a product.

02:15 In the original, in the one given, it is a reactant for what needs to be a product.

02:23 So it's multiply by negative one.

02:25 So what i like to do is i like to rewrite the whole equation using everything in the right order because it helps me do the canceling out when i add together all of the reactants.

02:49 So it's now 3h2 gas...

Please give Ace some feedback