# Question 0174d

Aug 27, 2015

#### Answer:

$\Delta {H}_{\text{sol" = -"45.7 kJ/mol}}$

#### Explanation:

I will assume that the problem wants you to determine the enthalpy change of solution of sodium hydroxide.

The enthalpy change of solution tells you what the change in enthalpy is when one mole of a substance is dissolved completely in a solvent (at constant pressure).

Since no mention was made about the specific heat of the resulting sodium hydroxide solution, you can safely assume it to be equal to that of pure water, 4.18"J"/("g" ^@"C").

So, you know that when you dissolve that much sodium hydroxide in 100.0g of water you get a temperature increase of

$\Delta T = {47.4}^{\circ} \text{C" - 23.6^@"C" = 23.8^@"C}$

The heat absorbed by the solution will thus be equal to

$q = {m}_{\text{sol" * c * DeltaT" }}$, where

$m$ - the mass of the solution;
$c$ - the specific heat of water;

Plug in your values to get

$q = \left(100.0 + 9.55\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 23.8color(red)(cancel(color(black)(""^@"C}}}}$

$q = + \text{10898.4 J" = +"10.9 kJ}$

If dissolving sodium hydroxide in water causes the temperature of the solution to increase, then that can only mean that the dissolution of sodium hydroxide releases heat, i.e. it is an exothermic process.

If you assume that no heat was lost to the exterior, then it follows that the amount of heat absorbed by the solution is equal to the amount of heat released by the dissolution of sodium hydroxide.

Keep in mind that you have to change the sign from poisitive to negative, since you're now expressing heat given off

${q}_{\text{released" = -q_"absorbed}}$

${q}_{\text{released" = -"10.9 kJ}}$

Now all you have to do is work out how many moles of sodium hydroxide produced this much heat. Use the compound's molar mass to get

9.55color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.2386 moles NaOH"

The enthalpy change of solution for sodium hydroxide will thus be

$\Delta {H}_{\text{sol" = q_"released"/"no. of moles}}$

$\Delta {H}_{\text{sol" = (-"10.9 kJ")/"0.2386 moles" = -"45.683 kJ/mol}}$

Rounded to three sig figs, the number of sig figs you have for the mass of sodium hydroxide, the answer will be

DeltaH_"sol" = color(green)(-"45.7 kJ/mol")#