Question #b8917

1 Answer
Aug 29, 2015

The answer should actually be #"1825 J"# and #-"228 kJ mol"^(-1)#.



In an experiment to determine the enthalpy of reaction, 100 mL of 0.0800 M #"CuSO"_4# solution was reacted with excess zinc metal in a calorimeter (total heat capacity including solution is #"500 J K"^(-1)#).

#"Cu"_text((aq])^(2+) + "Zn"_text((s]) -> "Cu"_text((s]) + "Zn"_text((aq])^(2+)#

The temperature of the solution in the calorimeter rose by a total of 3.65 K. The heat (q) absorbed by the calorimeter, the enthalpy change, #DeltaH#, for the reaction are respectively:

(a)182.5 J, +2281 kJ mol-1 (b) 1825 J, 228 kJ mol-1 (c) 1825 J, 22.8 kJ mol-1 (d) 1825 J, +228 kJ mol-1 (e) 1825 J, 228 J mol-


So, you know that you're dealing with the reaction between zinc metal and a solution of copper sulfate.

Heat capacity is simply a measure of how the amount of heat added to or removed from a substance and the temperature change that accompanies this process are related.

More specifically, heat capacity tells you what the ratio between added/removed heat and change in temperature.

In your case, you know that the total heat capacity of the system, which includes that of the solution, is #"500 J K"^(-1)#. This tells you that in order to increase the temperature of the system by #"1 K"#, you need to provide it with #"500 J"# worth of heat.

Since the temperature of the solution increased by #"3.65 K"#, the amount of heat absorbed must have been

#3.65color(red)(cancel(color(black)("K"))) * "500 J"/(1color(red)(cancel(color(black)("K")))) = color(green)("1825 J")#

If the temperature of the solution increased, then that can only mean that the reaction was exothermic, i.e. it released heat.

Assuming that no heat was lost to the exterior, the amount of heat absorbed by the solution + calorimeter must be equal to the amount of heat given off by the reaction.

This means that you have

#q_"sys" = - n * DeltaH" "#, where

#n# - the number of moles of copper sulfate that take part in the reaction.
#DeltaH# - the enthalpy change of reaction per mole.

Use the solution's volume and molarity to determine how many moles of copper sulfate were present

#color(blue)(C = n/V implies n = C * V)#

#n = "0.08 M" * 100 * 10^(-3)"L" = "0.008 moles CuSO"""_4#

So if the reaction of 0.008 moles of copper sulfate realeased 1825 J of heat, then the enthalpy change per mole will be

#DeltaH = (-"1825 J")/"0.008 moles" = -"228125 J/mol"#

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify one sig fig for both results

#DeltaH = color(green)(-"228 kJ mol"^(-1))#