1. AS Chemistry CALORIMETRY IN SOLUTION

2. MEASURING ENTHALPY CHANGES,  H, IN SOLUTION 1. Reaction carried out in insulated calorimeter eg polystyrene cup 2. Known volume (V) and concentration (C) or known mass (m /g) ( + known volume of water) of reagent 1 added 4. Stir and monitor the temperature of the solution at known times until steady 5. Add excess of reagent 2 to start reaction and ensure complete reaction 6. Stir continuously and monitor the temperature of the reaction mixture at known times until maximum or minimum reached and for several readings after that 7. Plot T against time to find ΔT Number moles used = n = C x V dm3 Temperature of water INCREASES if reaction is exothermic ; DECREASES if reaction is endothermic or number moles used = n = m / M r

3. T / ºC t / s Start of reaction ΔTΔT T / ºC t / s Start of reaction ΔTΔT Measuring  T Extrapolate graph back to start of reaction Measure ΔT as shown (ie ΔT that would occur if the reaction was instantaneous with no heat loss) EXOTHERMIC ENDOTHERMIC Heat change during reaction (/J) = q = m  C   T Calculating  H = energy (in JOULES) needed to warm ONE GRAM of water by 1K specific heat capacity ( / J K -1 g -1 ) of water C =  H = ± q / n J mole -1 ΔH + if endothermic ΔH - if exothermic Such experiments ALWAYS involve 2 major errors : 1. Heat loss2. Incomplete reaction mass of water in mixt. (/g) m = number of moles reacted n =

4. DATA VALUE Initial temperature of "X" 20.5 o C Initial temperature of "Y" 19.5 o C Final temperature of reaction mixture 9.8 o C Specific heat capacity of water 4.18 Jg -1 K -1 100 cm 3 of 0.500M aq. "X" is measured into a calorimeter and it's temperature recorded to 0.1 o C. Similarly, 100 cm 3 of 1.000M aq. "Y" (an excess) is measured and it's temperature recorded to 0.1 o C. "Y" is then added to "X" in the calorimeter, the reaction mixture is stirred and the maximum or minimum temperature is recorded after correcting for heat loss. m = 100 + 100 = 300 g  T = 9.8 – (20.5 + 19.5)/2 n(X) = CV = 0.500 x 100/1000 = - 10.2°C = 0.0500 moles Heat change during reaction = q = m  C   T = 200  4.18  10.2 = 8527 J = 8.53 kJ (3sf)  H = + q / no. moles = + 8.53 / 0.0500 = + 171 kJ mole -1 of X  T –ve   H +ve

5. The End