# Question ade0c

Aug 3, 2016

Here's what I got.

#### Explanation:

The idea here is that the heat given off by your reaction will be absorbed by the calorimeter.

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

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

So, you know that the temperature of the calorimeter increased by ${3}^{\circ} \text{C}$, since

$\Delta T = {28}^{\circ} \text{C" - 25^@"C" = 3^@"C}$

Notice that the calorimeter has a heat capacity of ${\text{1 kJ K}}^{- 1}$. As you know, changes in temperature are identical for temperature expressed in degrees Celsius and for Kelvin, you can say that the calorimeter has a heat capacity of ${\text{1 kJ"^@"C}}^{- 1}$.

Now, heat capacity tells you how much heat must be added or removed from a system in order to cause a ${1}^{\circ} \text{C}$ change in its temperature.

In your case, to cause the temperature of the calorimeter to increase by ${1}^{\circ} \text{C}$ you need to provide it with $\text{1 kJ}$ of heat. Since the temperature of the calorimeter increased ${3}^{\circ} \text{C}$, you know that it must hav absorbed

3 color(red)(cancel(color(black)(""^@"C"))) * overbrace("1 kJ"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("heat capacity")) = "3 kJ"

Now focus on the chemical reaction given to you. You know that you're dealing with the synthesis of calcium fluoride, ${\text{CaF}}_{2}$, from its constituent elements in their stable state

${\text{Ca"_ ((s)) + "F"_ (2(g)) -> "CaF}}_{2 \left(s\right)}$

The standard enthalpy change of formation, $\Delta {H}_{f}^{\circ}$, tells you the enthalpy change that accompanies the formation of one mole of a compound from its constituent elements in their stable state under standard conditions.

Use calcium metal's molar mass to convert the sample to moles

1 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078 color(red)(cancel(color(black)("g")))) = "0.02495 moles Ca"

Fluorine gas is said to be in excess, which means that all the moles of calcium metal will take part in the reaction. As a result, your reaction will produce -- keep in mind that you have a $1 : 1$ mole ratio between calcium and calcium fluoride in the balanced chemical equation

0.02495 color(red)(cancel(color(black)("mole Ca"))) * "1 mole CaF"_2/(1color(red)(cancel(color(black)("mole Ca")))) = "0.02495 moles CaF"_2

So, you know that when $0.02495$ moles of calcium fluoride are formed, the reaction gives off $\text{3 kJ}$ of heat. This means that when $1$ mole of calcium fluoride is formed, the reaction will give off

1 color(red)(cancel(color(black)("mole CaF"_2))) * "3 kJ"/(0.02495color(red)(cancel(color(black)("moles CaF"_2)))) = "120.24 kJ"#

Therefore, the standard enthalpy change of formation for calcium fluoride will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\Delta {H}_{f}^{\circ} = - {\text{120 kJ mol}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs.

SIDE NOTE The actual standard enthalpy of formation for calcium fluoride is

$\Delta {H}_{f}^{\circ} = - {\text{1219.6 kJ mol}}^{- 1}$

As you can see, the answer is about one order of magnitude off, so it's worth taking the time to double-check the values given to you.