Average bond enthalpies are generally defined for gasphase...

Average bond enthalpies are generally defined for gasphase molecules. Many substances are liquids in their standard state. $\mathrm{OB}$ (Section 5.7) By using appropriate thermochemical data from Appendix C, calculate average bond enthalpies in the liquid state for the following bonds, and compare these values to the gas-phase values given in Table 8.4: (a) $\mathrm{Br}-\mathrm{Br}$, from $\mathrm{Br}_{2}(l) ;$ (b) $\mathrm{C}-\mathrm{Cl}$, from $\mathrm{CCl}_{4}(l) ;$ (c) $\mathrm{O}-\mathrm{O}$, from $\mathrm{H}_{2} \mathrm{O}_{2}(l)$ (assume that the $\mathrm{O}-\mathrm{H}$ bond enthalpy is the same as in the gas phase). (d) What can you conclude about the process of breaking bonds in the liquid as compared to the gas phase? Explain the difference in the $\Delta H$ values between the two phases.

Key Concepts

Standard States and Physical State Corrections

Standard state refers to the physical form (solid, liquid, or gas) of a substance under standard conditions (usually 1 atm and a specified temperature). When calculating bond enthalpies for substances not in the gas state, corrections must be made to account for the differences in intermolecular interactions and the enthalpy changes associated with phase transitions, ensuring an accurate comparison of bond strengths across different physical states.

Phase Dependence of Bond Enthalpies

Bond enthalpies are generally defined for gas-phase species because the absence of intermolecular interactions in the gas phase provides a clearer measure of the intrinsic bond strength. In contrast, in the liquid phase, additional interactions, such as intermolecular forces and solvation effects, influence the energy required to break bonds, often leading to different enthalpy values compared to the gas phase.

Hess's Law and Thermochemical Cycles

Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the pathway taken. This principle is used in thermochemical cycles to calculate bond enthalpies by combining standard enthalpies of formation for reactants and products, thereby enabling estimation of bond energies even when direct measurements are not feasible.

Average Bond Enthalpy

Average bond enthalpy refers to the mean energy required to break a particular type of bond across a variety of compounds. Since bond energies can vary depending on the molecular environment, average values are often calculated from experimental data to provide a generalized measure.

Bond Enthalpy

Bond enthalpy is the energy required to break one mole of a bond in a gaseous molecule under standard conditions. It represents the strength of a chemical bond and serves as a basis for understanding chemical reactivity and stability.

Transcript

00:01 We are given the task of finding average bond entropies for substances in the liquid state and when the liquid state is the standard state.

00:13 We're going to be using thermochemical data from an appendix in the book to find those average bond enthalpys for liquid states for three different liquids.

00:25 One is br2 liquid and that's going to give us two br gas one is i'll just do one at a time okay so when we're doing this this one actually is pretty easy we look up our values of course this is zero and br as a gas is 223 or excuse me 118 .6 kilojoules per mole.

01:12 And remember to find our enthalpy for the reaction, it's always the sum of the delta h for the products minus the sum of the delta h of the reactants.

01:40 Okay, so this one is very simple.

01:52 It's going to be two times 118.

02:03 Wrong.

02:06 111 .8.

02:07 111 .8.

02:11 111 .8.

02:12 And again i just look these values up in the back of the book.

02:16 Minus 0 equals 2 .23 .6 kilojoules.

02:27 And it'll be per mole of br2.

02:31 Okay.

02:33 So when i look this up in table 84, we are given the table value for br br is 193 kilojoules per mole.

03:05 So what we see is that our calculated is greater than the table value.

03:19 Let's do another one.

03:24 How about a brown? b is carbon tetrachloride.

03:33 And that is going to work like this.

03:41 And please note that there are four.

03:42 Whoops that should be a g there are four ccl bonds so we've got c and then one two three four so we've got four ccl bonds and we are tasked with finding the ccl bond for this liquid we're going to calculate the enthalpy of the reaction just like we did this last one i'm going to write the values right here 718 erase that 7 and start again 718 point for 121 .7.

04:34 I'm just looking these up and negative 139 .3.

04:43 So our heat for the reaction is going to equal 1 times 718 .4, that's kilgels per mole, plus 4 times 121 .7 kilgible's per mole, minus a negative 139 .3.

05:12 Also kilojoules per mole.

05:15 When you do the math on this, i got 1 ,305.

05:28 A nicer 5 there, kilojoules.

05:38 And then that is going to be per 4 c, cl bonds.

05:45 So i have to divide that by 4.

05:47 And i got 338.

05:52 I can't read my decimal point there, so i'm just going to go 338 kilojoules per mole, and our table value for this one is 328 kilojoules per mole.

06:15 So again, when we compare these two, this statement is still true.

06:20 Our calculated value is greater than the table value.

06:25 Last but not least, we are tasked with finding the o -o bond in h -2 -o -2.

06:35 And this was a little more complicated.

06:37 I hope i did this one correctly.

06:52 So this is the first thing we're going to do.

06:56 And this will give us the heat of the reaction...