Question #ade0c

1 Answer
Aug 3, 2016

Here's what I got.


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

#-q_"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^@"C"#, since

#DeltaT = 28^@"C" - 25^@"C" = 3^@"C"#

Notice that the calorimeter has a heat capacity of #"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 #"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^@"C"# change in its temperature.

In your case, to cause the temperature of the calorimeter to increase by #1^@"C"# you need to provide it with #"1 kJ"# of heat. Since the temperature of the calorimeter increased #3^@"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, #"CaF"_2#, from its constituent elements in their stable state

#"Ca"_ ((s)) + "F"_ (2(g)) -> "CaF"_ (2(s))#

The standard enthalpy change of formation, #DeltaH_f^@#, 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 #"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

#color(green)(|bar(ul(color(white)(a/a)color(black)(DeltaH_f^@ = - "120 kJ mol"^(-1))color(white)(a/a)|)))#

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

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

#DeltaH_f^@ = -"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.