ΔHn=(2361.7 J)(1 KJ/ 1000 J)/(1.25 mol H2O) = 1.89 KJ/mol H2O. In trial 2 it is ΔHn=(2445.3 J)(1 KJ/ 1000 J)/(1.25 mol H2O) = 1.96 KJ/mol H2O. Since there are two trials, the average was taken by adding the enthalpies together and then dividing it by 2, (1.89 KJ/mol H2O + 1.96 KJ/mol H2O)/2 = 1.92 KJ/mol H2O. 0:001:122:243:364:486:002022242628303234Trial 1 (HNO₃)Trial 2 (HNO₃)Time (minutes and seconds)Temperature (°C)Graph 3. Temperature vs. Time of Neutralization of NaOH and HNO3: The graph above shows the temperature of acid after base was added into the Calorimeter. At time 0:00, is when the reaction of NaOH and the HNO3started. The Black line was used to represent a makeshift trend line to determine the maximum temperature. The blue diamondsrepresent Trial 1 and the orange squares represent Trial 2.In Graph 3, above, the black line represents the trend line that determines the maximum temperature of the reaction. In Trial 1 the maximum temperature is 30.2 °C, and in Trial 2 it is 27.0 °C. The exact molar concentration of the HNO3was 1.1 M, and the molar concentration of the NaOH was 1.0 M. In the reaction, 50 mL of base and 50 mL of acid were used.Averageinitial temp.of HCl andNaOH (°C)Maximum temp.(°C)Temp.change(°C)Volumeof finalmixture(mL)Mass offinalmixture(g)Heatevolved(J)Moles ofOH⁻reacted(mol)Moles ofH2Oformed(mol)Enthalpy ofNeut. (KJ/molH2O)Trial 121.1 °C30.2 °C9.1 °C100 mL100 g3712.8 J1.25 mol1.25 mol2.97 KJ/mol H2OTrial 221.65 °C27.0 °C5.35°C100 mL100 g2236.3 J1.25 mol1.25 mol1.79 KJ/mol H2OTable 3. Enthalpy of Neutralization of NaOH and HNO3: The graph above shows the results for Part II.2. The mass of the final mixture was gathered by adding the volume of the base and acid solutions, assuming the density is the dame as water. The maximum temperature was obtained from Graph 3. The Heat evolved was calculated by Equation1, and Equation 4 was used to calculate the enthalpy of neutralization.
Table 3, above, shows the results of Part II.2, the neutralization of NaOH and HNO3. Equation 1 was used to first find the heat evolved by the reaction. This is represented by q=(100 g)(4.18 J/g °C)(9.1 °C) = 3712.8 J in trial 1, and q=(100 g)(4.18 J/g °C)(5.35 °C) = 2236.3 J in trial 2. This is then divided by the moles of water produced as stated in Equation 4, but make sure to convert it to KJ instead on leaving it in J. In trial 1 the enthalpy of neutralization is ΔHn=(3712.8 J)(1 KJ/ 1000 J)/(1.25 mol H2O) = 2.97 KJ/mol H2O. In trial 2 it is ΔHn=(2236.3 J)(1 KJ/ 1000 J)/(1.25 mol H2O) = 1.79 KJ/mol H2O. Since there are two trials, the average was taken by adding the enthalpies together and then dividing it by 2, (2.97 KJ/mol H2O + 1.79 KJ/mol H2O)/2 = 2.38 KJ/mol H2O. Discussion:Before the experiment was conducted, it was hypothesized that the specific heat of the unknown metal would be approximately 0.5 J/g °C. I also hypothesized that the enthalpy of neutralization will be ~1.0-1.5 KJ/mol H2O. The specific heat of the unknown metal was 0.46 J/g °C which confirms my hypothesis. The enthalpy of neutralization of HCl and HNO3were 1.92 KJ/mol H2O, and 2.38 KJ/mol H2O which is higher than I anticipated.
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