# Question f776d

Dec 8, 2017

$- {\text{57.1 kJ mol}}^{- 1}$

#### 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.
.

${\text{H"_ 2"SO"_ (4(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

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

${V}_{\text{total" = "60.0 mL" + "60.0 mL}}$

${V}_{\text{total" = "120.0 mL}}$

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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

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
• $\Delta T$ is the change in temperature, equal to the difference between the final temperature and the initial temperature of the solution

$\Delta T = {26.75}^{\circ} \text{C" - 24.09^@"C}$

$\Delta T = {2.66}^{\circ} \text{C}$

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")))

${q}_{\text{absorbed" = "1335.5 J}}$

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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{q}_{\text{given off" = -q_"absorbed}}}}}$

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

This means that you have

$\Delta {H}_{\text{rxn" = - q_"absorved}}$

$\Delta {H}_{\text{rxn" = -"1335.5 J}}$

This tells you that when $0.0234$ moles of water are produced, $\text{1335.5 J}$ of heat are being given off $\to$ 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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{neutralization" = -"57.1 kJ mol}}^{- 1}}}}$