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25 cm3 of 1 M lead(II) nitrate solution was added to 50 cm3 of 1 M hydrochloric acid and a precipitate was formed. The original solutions had a temperature of 18.9 ºC which on mixing rose to 20.7 ºC. Write down full and ionic equations for the reaction and calculate the enthalpy change. What is limiting the accuracy of the result? Specific heat capacity of water = 4.18 J g-1 K-1.
|
Pb(NO3)2(aq) + 2HCl(aq) ⇒ PbCl2(s) + 2HNO3(aq) |
|
Pb2+(aq) + 2Cl-(aq) ⇒ PbCl2(s) |
There is a total of 50 + 25 = 75 g of "water" present
Temperature rise = 20.7 - 18.9 = 1.8 ºC
\ heat released = 4.18 × 75 × 1.8 = 564 J = 0.564 kJ
As the solution concentrations are the same and there is twice the volume of hydrochloric acid, we must have twice as many moles of the acid. This is in the same ratio as the equation, so should react exactly. Always give your enthalpy value for the number of moles shown in the equation. In this case, we want the number of joules per mole of lead(II) nitrate or per two moles of hydrochloric acid. So the number of moles to use here is the number of moles of lead(II) nitrate:
Number of moles of Pb(NO3)2 = 1 × 25/1000 = 0.025 mol
\ enthalpy change = - 0.564/0.025 = - 23 kJ mol-1
Note the negative sign for the exothermic reaction and the use of two sig. figs. Although the concentration was only stated to 1 sig. fig. (1 M), it is highly unlikely that it would not be made up much more accurately than this. The temperature change is only accurate to 2 sig. figs. (1.8 ºC), and it is this which is limiting the accuracy. We could improve our result either by obtaining a higher temperature rise (with more concentrated solutions) or by using a more accurate thermometer.