Measuring cylinders, thermometer and plastic cup
0.5 mol dm-3 silver nitrate solution, 0.5 mol dm-3 sodium chloride solution
1. 25cm3 of 0.5 mol dm-3 sodium chloride solution is measured and poured into a clean and dry plastic cup using a measuring cylinder.
2. The initial temperature of sodium chloride solution is measured and then recorded.
3. Using another measuring cylinder, 25 cm3 of 0.5 mol dm-3 silver nitrate solution is measured. The initial temperature of the silver nitrate solution is measured after 5 minutes and recorded.
4. The silver nitrate solution is poured quickly and carefully into the sodium chloride solution.
5. The mixture is stirred with a thermometer throughout the experiment and the highest temperature obtained is recorded.
= (25 + 25) g = 50 g
Initial temperature of sodium chloride solution = T1oC
Initial temperature of the silver nitrate = T2 oC
Highest temperature reached = T4 oC
=> Average initial temperature = (T1+T2)/2 = T3 oC
è Increase in temperature, q = (T4 - T3) oC
No of moles of sodium chloride used
No of moles of silver nitrate used
= MV/1000 = 0.0125 mole
= MV/1000= 0.0125 mole
Chemical equation: AgNO3 (aq) + NaCl (aq) à AgCl (s) + NaNO3 (aq)
Ionic equation: Ag + (aq) + Cl- (aq) à AgCl (s)
1 mole 1 mole 1 mole
Heat released = mcq
= (25 + 25 ) x 4.2 x (T4 - T3)
= x J
0.0125 mole of AgCl precipitated releases x J of heat energy.
Therefore, 1 mole of AgCl precipitated will released
Heat of precipitation, DH = x/0.0125 J = y J
AgNO3 (aq) + NaCl (aq) à AgCl (s) + NaNO3 (aq) DH = - y/1000 kJmol-1