Question #d0e3c

The idea here is that the enthalpy change of reaction, `DeltaH_"rxn"`, for this neutralization reaction caused the temperature of the resulting solution to increase from `22.21^@"C"` to `26.84^@"C"`.

This of course implies that the neutralization reaction was exothermic, i.e .it gave off heat. As a result, the value of `DeltaH_"rxn"` must be negative, since a minus sign is used to designate heat lost.

Now, barium hydroxide, `"Ba"("OH")_2`, will react with hydrochloric acid, `"HCl"`, to produce aqueous barium chloride, `"BaCl"_2`, and water, according to the balanced chemical equation

`"Ba"("OH")_ (2(aq)) + color(red)(2)"HCl"_ ((aq)) -> "BaCl"_ (2(aq)) + color(blue)(2)"H"_ 2"O"_((l))`

Notice that you have a `1:color(red)(2)` mole ratio between barium hydroxide and hydrochloric acid. This means that the reaction consumes twice as many moles of hydrochloric acid than moles of barium hydroxide.

Use the molarities and volumes of the two solutions to determine how many moles of each you're mixing

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))`

Plug in your values to get

`n_(Ba(OH)_2) = "0.340 M" * 60.0 * 10^(-3)"L" = "0.0204 moles Ba"("OH")_2`

`n_(HCl) = "0.680 M" * 60.0 * 10^(-3)"L" = "0.0408 moles HCl"`

As you can see, you have exactly the number of moles of acid you need in order to make sure that both reactants are completely consumed by the reaction, i.e. you're not dealing with a limiting reagent.

So, pick one of the two reactants and use the mole ratio that it has with water to determine how many moles of water you get here.

Since hydrochloric acid and water have a `color(red)(2):color(blue)(2)` mole ratio, it follows that the reaction will produce the same number of moles of water as you have moles of hydrochloric acid that are taking part in the reaction.

This means that you have

`n_(H_2O) = "0.0408 moles H"_2"O"`

Use water's molar mass to determine how many grams would contain this many moles

`0.0408 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.735 g"`

Now, to calculate the heat absorbed by the solution, use the equation

`color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "`, where

`q` - the amount of heat gained / lost
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature

You can approximate the specific heat of the solution to be equal to that of water

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c_"water" = "4.18 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))`

In your case, the change in temperature is

`DeltaT = 26.84^@"C" - 22.21^@"C" = 4.63^@"C"`

Assuming that the resulting solution has a density of `"1.00 g mL"^(-1)`, you can say that its total mass will be

`(60.0 + 60.0)color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "120.0 g"`

Plug in your values to get

`q = 120.0 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 4.63color(red)(cancel(color(black)(""^@"C"))) = "2322.3 J"`

So, if this is how much heat was absorbed by the solution, you can say that this is how much heat was released by the neutralization reaction.

Since this much heat corresponds to the formation of `0.0408` moles of water, it follows that one mole of water would require

`1 color(red)(cancel(color(black)("mole H"_2"O"))) * "2322.4 J"/(0.0408color(red)(cancel(color(black)("moles H"_2"O")))) = "56,922 J"`

Expressed in kilojoules and rounded to three sig figs, the enthalpy change of reaction per mole of water produced will thus be

`DeltaH_"rxn" = color(green)(|bar(ul(color(white)(a/a)-"56.9 kJ/mol H"_2"O"color(white)(a/a)|)))`