Was quickly added to the contents of the calorimeter

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was quickly added to the contents of the calorimeter – the resulting temperature of which was recorded immediately. For the next 4 minutes, the temperature was read and documented at 15 second intervals. The calorimeter was then emptied and cleaned. The procedure was repeated, this time by adding 50mL of a strong acid (1.000M HCl) to 50mL of a strong base (0.900M NaOH). A final trial was performed my mixing 50mL of dichloroacetic acid (CHCl2COOH) with 0.900M NaOH.CalculationsThe results of the temperature change subsequent to the addition of hot water to cold water are demonstrated in the graph below:
05010015020025030005101520253035f(x) = − 0 x + 30.76R² = 0.82Time (seconds)Temperature (°CFigure 1. This graph represents the change in temperature of a solution where cold (22.4°C)water was mixed with hot water (42.4°C). The temperature was measured at 15 second intervals. The extrapolation shows that the temperature at the instant of mixing in the calorimeter is 30.755°C.Heat Capacity of Calorimeter:|Heat lost by H2O| = | Heat gained by H2O | + | Heat gained by the calorimeter|∆T = │Tfinal – Tinitial (hot)∆T = │30.5°C– 42.4°C │ ∆T = │303.65K– 315.55K │ ∆T = 11.9K (50mL 1g/mL 4.184J/gK 11.9K) = (50mL 1g/mL 4.184J/gK 8.1K) + (Cpcal 2489.48J = 1694.52J + (Cpcal 8.1K) Cpcal= 98.1432 J/KTherefore the heat capacity of the calorimeter is 98.1432 J/K.∆T’ = │ Tfinal – Tinitial (cold)∆T’ = │30.5°C– 22.4°C │ ∆T’ = │303.65K– 295.55K │ ∆T’ = 8.1K
8.1K)
Enthalpy of neutralizationThe results of the temperature change subsequent to the addition of 1.000M HCl to 0.900M NaOH in a calorimeter are demonstrated in the graph below:050100150200250300051015202530f(x) = − 0 x + 28.6R² = 0.86Time (seconds)Temperature (°C)Figure 2. This graph above represents the change in temperature occurring during the neutralization process between a strong acid (HCl) and a strong base (NaOH). The extrapolation shows that the temperature at the instant of mixing in the calorimeter is 28.599°C.Calculation of qpqp= Cp∆Tqp= 98.1432J/K (28.599°C – 28.2°C) ---- essentially it is unnecessary to change °C to Kelvin when calculating ∆T.qp= 39.2572 J∆T = 0.4°CCalculation of number of moles of acid and base neutralized in each case Since it was revealed in the course outline that the base was completely consumed in the reaction, this means that the solution at equilibrium must remain slightly acidic 5 ; pH<7. Therefore we can assume that the moles of base added is equal to the moles of H3O+neutralized: 0.900M 0.050L = 0.045 moles of base.| Heat released by reaction | = | Heat gained by acid + base + calorimeter |H3O++ OH-2H2O ---------- 0.045 moles of H3O+

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