 [SPM Chemistry Thermochemistry] Form 5 chapter three

Hi there,

SPM Chemistry Thermochemistry Heat of Precipitation Question

1. Heat released during the precipitation reaction between 200 \mathrm{~cm}^{3} of lead (II) nitrate solution and 200 \mathrm{~cm}^{3} of potassium sulphate solution raises the temperature of the mixture solution by 6^{\circ} \mathrm{C}. Calculate the amount of heat released. (specific heat capacity of the solution = 4.2 \mathrm{Jg}^{-1}{ }^{\circ} \mathrm{C}^{-1})

SPM Chemistry Thermochemistry Heat of Precipitation Solution

The energy released from the heat of precipitation, can be related to the heat absorbed by the solution as shown in the following formula, where

Q = amount of heat released (J)
m = total mass of the solution (g)
cp = specific heat capacity of solution (4.2 J/g°C)
ΔT = the temperature change which is 6 °C

Q = mcpΔT

Assuming that the density of the solution is similar to water, ρ = 1000 kg/m3 = 1 g/cm3
Using a total volume of 400 cm3, this corresponds to:

m = 400 cm3 x 1 g/cm3 = 400 g

Using all the information we have, assuming NO HEAT LOSS TO THE SURROUNDINGS,

Q = 400(4.2)(6) = 10 080 J = 10.08 kJ

In this case, you do not need the stoichiometry to find out the heat released since temperature change and mass of solution can be determined.

Thank u so much …  