Thermochemistry and calorimetry

Question

If 80.5 g of KBr are dissolved in 400.0 g of water at 25.0 °C in an insulated container, a temperature change is observed. The ∆H of solution of KBr is 19.9 kJ/mol. Assuming that the specific heat of the solution is 4.184 J/(g°C), and that no heat is gained or lost by the container, what will be the final temperature of the solution?

Answer 1

For calorimetry of a reaction, the heat given off by the reaction heats the solution and the calorimeter:

-n_i(ΔΗ_rxn./ν_i) = m_sC_sΔT + C_calΔT

i indicates the critical species (in this case, KBr dissolving). v_i is the stoichiometric coefficient of the critical species in the reaction statement that corresponds to the ΔH.

s is indicating that we have to use properties of the solution (You must use the mass of the salt and the water!). Usually, the solution is considered dilute enough that you can use c_w as done here. The assumption is made that the calorimeter has no effect so the third term is out.

The rxn statement is simply KBr(s) --> K⁺(aq) + Br^-(aq) (stoichiometry is all ones)

The energy released (a heat of solution) :

-80.5 g * (1mole/119.0 g)(19900 J/mole) (note that I have assumed the problem is giving a positive heat of solution for KBr dissolving in water) (This will lower the temperature and ΔT will be less than 0

Now solve by plugging in the value of energy absorbed and make it equal to

(80.5g + 400 g) * (4.184 J/g°C) ΔT Solve for the difference in temperature and then add it to 25° to find the initial temperature.

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