What is the heat released in this reaction?

Question

A reaction known to release 1.78 kj of heat takes place in a calorimeter containing 0.100 L of solution. The temperature rose by 3.65 C. Next, a small piece of calcium carbonate was put in the same container AND 0.100 L of dilute HCl was poured over it. The temperature of the calorimeter rose by 3.57 C. What is the heat released?

Why does the question include stuff like calcium carbonate? Is that going to decompose to CO2 and CaO? and how does more than one reaction taking place affect the answer? How should I go about solving this?

Would I make delta T 3.57? or 7.22? also, would the mass be 100 grams? or would it include the original 0.100 L of solution? thanks once again

Answer 1

The first part of the question gives you the information you need to determine the "calorimeter constant", or the amount of heat required to change the temperature of the whole calorimeter set-up by 1 degree.

The calcium carbonate/HCl stuff is the reaction whose enthalpy change you are trying to determine, using the same calorimeter as before. When you combine calcium carbonate and HCl, you'll get a reaction that produces CO2 and leaves you with a solution of calcium chloride. So the overall reaction that you're looking at is:

CaCO3 + 2 HCl --> CO2(g) + H2O + CaCl2

Calcium carbonate will only decompose to CO2 and CaO when you heat it--not when you react it with an acid.

Hope this helps....I'll let you do the calculations.