Lab 10_Heat of Reaction for the Neutralization of Hydrochloric Acid With Sodium Hydroxide Solution

Alex T. Zhao Gregg Kerr IB Chemistry HL February 23, 2013 Heat of reaction for the neutralization of hydrochloric acid with sodium hydroxide solution Objective: To experimentally determine the molar heat of the neutralization reaction between Hydrochloric Acid and Sodium Hydroxide by measuring the temperature increase that occurs. Equipment:  Polystyrene cup  Graduated cylinder (100 mL)  Thermometer (-10-100°C)  Hydrochloric Acid (HCl) 1 mol dm-3 (50 mL)  Sodium hydroxide solution (NaOH) 1 mol dm-3 (50 mL) Procedure: 1. Place 50 mL of 1 mol dm-3 HCL into the polystyrene cup and measure its temperature. 2. Measure 50 mL of 1 mol dm-3 of NaOH into a measuring cylinder and record its temperature. 3. Add the NaOh into the HCL, stir briefly with the thermometer, and note the maximum temperature reached. Data collection: Volume of HCl (mL ± 1mL) 50 Volume of NaOH (mL ± 1mL) 50 Int. Temp HCl (°C ± 0.2°C) 22.0 Int. Temp NaOH (°C ± 0.2°C) 21.8 Average Temp* (°C ± 0.2°C) (22.0+21.8)/2=21.9 Final Temp (°C ± 0.2°C) 26.0 *Average Temp will be used as the Initial temperature of the system, denoted by Ti Processing of results, and questions 1. Using the average temperature of the two reactants as the initial temperature, calculate the change in temperature of the solution:

∆ T  Tf  Ti  26.0C  21.9 C  4.1C 2. Calculate the amount of heat energy released Let “Q” be heat energy released; “m” be mass; “C” be specific heat capacity; ∆T be change in temperature, then:

IB Chemistry HL Lab 10

February 23, 1013

Q  m C  ∆ T Q  (0.100kg)  (4.18  10 3 J C 1Kg 1 )  (4.1C) Q  1713.8 J , Thus, 1713.8 joules of heat energy was released in the reaction. Since the heat was released, the change in enthalpy, ∆H, should be negative, hence ∆H1= -1713.8 J. 3. The answer to question 2 is the heat released when 50 mL of 1 mol dm-3 HCl is neutralized with 50 mL of 1 mol dm-3 NaOH a. Calculate the molar heat of reaction.

n CV n (1 mol dm 3 )  (0.050 dm 3 ) n 0.05 mol This means that for 0.005 mol of reactants, 1713.8 J of heat is released. Thus, using ratios, we can find the molar heat of reaction. n1 n  molar ∆ H1 ∆ H molar ∆ H molar 

nmolar  ∆ H1 n1

(1 mol)  (1713.8 J ) 0.05mol  34276 J  34.276KJ

∆ H molar  ∆ H molar

b. Is this reaction exothermic or endothermic? i. Exothermic c. Write the molar value of ∆H for the neutralization reaction using the appropriate sign. HCl(aq)  NaOH (aq)  NaCl(aq)  H 2O(l ) H  34.276KJ i. , or the molar heat can be written with the units KJ/mol, as -34.3 KJ/mol. Since the reaction is exothermic, the ∆H is negative Percentage Error: Valueexp  ValueTheo %Error   100% Valuetheo

(34.276KJ / mol)  (56.2KJ / mol)  100% 56.2KJ / mol %Error  39.0% %Error 

Conclusion: We have determined in this experiment that the neutralization of hydrochloric acid with sodium hydroxide is an exothermic reaction, with an IB Chemistry HL Lab 10

February 23, 1013

experimental molar heat of reaction of -34.3 KJ/mol. The theoretical molar heat of reaction is -56.2KJ/mol, which yields a percentage error of 39%. The greatest source of error was that the system was not completely isolated, hence heat was constantly lost to the surroundings. Due to this loss of heat, the temperature rise was not as great as expected, which caused the empirical molar heat of reaction to be lower than the theoretical molar heat of reaction.

IB Chemistry HL Lab 10

February 23, 1013