# Question 8e0a7

Nov 10, 2015

$+ \text{2.05 kJ/mol}$

#### Explanation:

The idea here is that you need to use the change in temperature and the heat capacity of the calorimeter to determine the enthalpy change of reaction, $\Delta {H}_{\text{rxn}}$, that corresponds to dissolving $\text{15.3 g}$ of sodium nitrate in water.

Once you know the value of $\Delta {H}_{\text{rxn}}$ for this reaction, you can use sodium nitrate's molar mass to help you find the enthalpy change of reaction for one mole of sodium nitrate.

The first important thing to notice here is that the temperature of the water decreases when the sample is dissolved. This means that the reaction is actually taking in heat from its surroundings, i.e. the water.

This tells you that the dissolution of sodium nitrate is an endothermic process, so you can expect $\Delta {H}_{\text{rxn}}$ to be positive.

For the calorimeter, the equation that establishes a relationship between heat lost and change in temperature looks like this

$\textcolor{b l u e}{q = C \cdot \Delta T} \text{ }$, where

$q$ - the heat lost by the water
$C$ - the heat capacity of the calorimeter
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

$q = 1071 \text{J"/color(red)(cancel(color(black)(""^@"C"))) * (21.56 - 25.00)color(red)(cancel(color(black)(""^@"C"))) = -"3684.2 J}$

The minus sign symbolizes the fact that the water is losing heat to the dissolution of the sodium nitrate. This means that you have

${q}_{\text{dissolution" = -q_"water}}$

The reaction absorbed

${q}_{\text{dissolution" = - (-"3684.2 J") = +"3684.2 J}}$

Now, use sodium nitrate's molar mass to help you find the number of moles you get in that $\text{15.2-g}$ sample

15.3color(red)(cancel(color(black)("g"))) * "1 mole NaNO"_3/(84.99color(red)(cancel(color(black)("g")))) = "0.18002 moles NaNO"_3

So, if the dissolution of $0.18002$ moles of sodium nitrate absorbed $\text{3684.2 J}$ of heat, it follows that th edissolution of one mole will absorb

1color(red)(cancel(color(black)("mole NaNO"_3))) * "3684.2 J"/(0.18002color(red)(cancel(color(black)("moles NaNO"_3)))) = "20465.5 J/mol"#

Rounded to three sig figs, the number of sig figs you have for the mass of sodium nitrate, and expressed in kilojoules per mole, the answer will be

$\Delta H = \textcolor{g r e e n}{+ \text{2.05 kJ/mol}}$