Question

Question #f776d

Answer 1

`-"57.1 kJ mol"^(-1)`

Explanation:

You know that you're dealing with a neutralization reaction that takes place between a strong acid and a strong base, so start by writing the balanced chemical equation.
.

`"H"_ 2"SO"_ (4(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l))`

Now, the reaction consumes sulfuric acid and sodium hydroxide in a `1:2` mole ratio, so use the molarities and the volumes of the two solutions to find how many moles of each reactant are present.

`60.0 color(red)(cancel(color(black)("mL"))) * ("0.810 moles H"_ 2"SO"_4)/(10^3color(red)(cancel(color(black)("mL")))) = "0.0486 moles H"_2"SO"_4`

`60.0 color(red)(cancel(color(black)("mL"))) * "0.390 moles NaOH"/(10^3color(red)(cancel(color(black)("mL")))) = "0.0234 moles NaOH"`

As you can see, the reaction requires `2` moles of sodium hydroxide for every `1` mole of sulfuric acid, so right from the start, you can say that sodium hydroxide will act as the limiting reagent here.

That is the case because `0.0234` moles of sodium hydroxide will only consume

`0.0234 color(red)(cancel(color(black)("moles NaOH"))) * ("1 mole H"_2"SO"_4)/(2color(red)(cancel(color(black)("moles NaOH")))) = "0.0117 moles H"_2"SO"_4`

The remaining moles of sulfuric acid will be in excess, i.e. they won't take part in the reaction. Moreover, since you have a `2:2` mole ratio between sodium hydroxide and water, you can say that the reaction will produce `0.0234` moles of water.

The total volume of the solution will be

`V_"total" = "60.0 mL" + "60.0 mL"`

`V_"total" = "120.0 mL"`

Since you can assume that this solution has the same density as water, you can say that the mass of the solution will be equal to

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

Next, use the fact that the heat absorbed by the solution can be calculated using the equation

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `q` is the heat absorbed by the solution
• `m` is the mass of the solution
• `c` is the specific heat of the solution, equal to that of water
• `DeltaT` is the change in temperature, equal to the difference between the final temperature and the initial temperature of the solution

In your case, you have

`DeltaT = 26.75^@"C" - 24.09^@"C"`

`DeltaT = 2.66^@"C"`

Plug your values into the equation to find the heat absorbed by the solution--keep in mind that the difference between two temperatures has the same value in degrees Celsius and in Kelvin!

`q_"absorbed" = 120.0 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)("K"^(-1)))) * 2.66 color(red)(cancel(color(black)("K")))`

`q_"absorbed" = "1335.5 J"`

Now, the idea here is that the heat absorbed by the solution is equal to the heat given off by the neutralization reaction.

`color(blue)(ul(color(black)(q_"given off" = -q_"absorbed")))`

The minus sign is used here because by definition, heat given off carries a negative sign.

This means that you have

`DeltaH_"rxn" = - q_"absorved"`

`DeltaH_"rxn" = -"1335.5 J"`

This tells you that when `0.0234` moles of water are produced, `"1335.5 J"` of heat are being given off `->` this essentially means that the neutralization reaction is exothermic.

You can thus say that when `1` mole of water is produced, the reaction gives off

`1 color(red)(cancel(color(black)("mole H"_2"O"))) * "1335.5 J"/(0.0234color(red)(cancel(color(black)("mole H"_2"O")))) = "57,072.6 J"`

This means that the enthalpy change of reaction when `1` mole of water is produced is equal to

`DeltaH_ ("rxn/mole H"_2"O") = -"57.1 kJ"`

The answer is rounded to three sig figs and expressed in kilojoules.

Alternatively, you can say that you have

`color(darkgreen)(ul(color(black)(DeltaH_"neutralization" = -"57.1 kJ mol"^(-1))))`