# Question #f776d

##### 1 Answer

#### 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 **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 **moles** of sodium hydroxide for every **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 **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 **mole ratio** between sodium hydroxide and water, you can say that the reaction will produce **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 themassof the solution#c# is thespecific heatof the solution, equal to that of water#DeltaT# is thechange in temperature, equal to the difference between thefinal temperatureand theinitial temperatureof 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 **moles** of water are produced, **given off** **exothermic**.

You can thus say that when **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 **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))))#