The Enthalpy of Neutralization of Phosphoric Acid Worksheet PDF

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Jiries Meehan-Atrash

Caitlin Bettenay

Wednesday 25th January 2017

The Enthalpy of Neutralization of Phosphoric Acid Worksheet Name: Caitlin Bettenay__ Date: 25/01/16____ Lab Section: Wednesday 5:30-8:30pm Describe coffee cup calorimetry and how it is used to find the enthalpy of various reactions that occur in aqueous solutions. Make sure to include the relevant equations. Why can you use the specific heat capacity and density of pure water to determine the enthalpy of reaction? What assumptions must be made in order to do this? A coffee cup calorimeter is a constant pressure calorimeter, thus it can be used to measure the heat which is equivalent to the change in enthalpy. The enthalpy, ΔH, of a system is the sum of the internal energy of the system and the product of pressure and volume. H is a state function. ΔH=E+PV. E = q+w, w= -PΔV, E = q+-PΔV. Where E = Energy, P = Pressure and V = volume. q = heat and w = work. During the experiment, all the heat evolved in an exothermic reaction is used to raise the temperature of a known mass of water. For endothermic reactions, the heat transferred from the water to the reaction can be calculated by measuring the lowering of temperature of a known mass of water. In order to calculate enthalpy, the following formula is used: q=cs × m × ∆ T Where q is the heat added, cs is the specific heat, and ∆ T is the change in temperature

Diagram 1: Graphical representation of a coffee calorimeter. m is the mass measured in grams

If the reaction is exothermic, then q is given a negative sign and if the reaction is edothermic, then a positive sign is given. The above equation represents the heat gained by the solution. In order to determine the heat of neutralization of phosphoric acid, the heat transfer associated with the reaction is needed ( q r ×n ). Assumptions: - That the specific heat capacity of the solution is equal to the specific heat capacity of the water. - That the constant figures of water density and specific heat capacity are constant - That 200ml of solution is equal to 200g as the density of water is 1 g / cm3 - That there is no heat that has been lost in the surroundings - Every calorimeter experiment must ensure that there is a focus on reducing the exchange of heat between the calorimeter contents and the surroundings.

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Jiries Meehan-Atrash

Caitlin Bettenay

Wednesday 25th January 2017

Data: Insert your data table (with a caption) below that captures all of the relevant information. Table 1: Table Displaying the Maximum and Minimum Temperature (C) for 2 Trials of Coffee Cup Calorimetry Conducted of 5 minute intervals. The Table Also Displays the Temperature Change (∆ T ) for Each Trial Including the Calculations Conducted for Trial 1. Trial 1:

Trial 2:

Maximum Temperature (C):

32.05C

32.24 C

Initial Temperature (C):

21.19 C

21.66 C

Temperature Change ∆T :

Temperature Change(∆ T )=Final Temperature ∴32.05 C−21.19 C=10.86 C

10.58C

Results: Report your calculated average value of ∆H of neutralization for phosphoric acid. Include hand written sample calculations. q=cs × m× ∆ T TRIAL 1: q=¿ change in heat. cs is the specific heat of water which is 4.18 J /( g ×C ) . m is the mass of the solution which is 100mL = 100g and the ∆ T for the first trial can be seen from table 1 which is 10.86 C .

q=4.18 J / (100 g ×C )× 100 g × 10.86 C q=4539.48 J∨4.53948 kJ TRIAL 2: q=¿ change in heat. cs is the specific heat of water which is 4.18 J /( g ×C ) . m is the mass of the solution which is 100mL = 100g and the ∆ T for the second trial can be seen from table 1 which is 10.58C . q=4.18 J / (100 g ×C )× 100 g × 10.58 C

q=4422.44 J ∨4.42244 kJ

Therefore, the average value of ∆H of neutralization for phosphoric acid is:

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Jiries Meehan-Atrash

Wednesday 25th January 2017

Caitlin Bettenay

Average enthalpy change=

( ∑ of trial1∧2 )

2 ( 4539.48 J + 4422.44 J ) Average enthalpy change= 2 Average enthalpy change=4480.96 J ∨ 4.48096 kJ This value is the heat gained by the water, but is the heat lost in the reaction between H3PO4 and NaOH causing an exothermic reaction, so the average change in enthalpy is −4480.96 J

Report the percent error for the ∆H of neutralization for phosphoric acid. Include any calculations. H3PO4 (l) + 3NaOH (l)  3H20 (l) + Na3PO4 (l) Table 2: Table displaying the standard enthalpy of formation ΔHf ° (kJ/mol) Standard Enthalpy of Formation ΔHf ° (kJ/mol) OH− (aq) −230 H+ (aq) 0 H2O (l) −286

Net ionic equation for the complete neutralization of H3PO4 with NaOH: 3 H+( aq)+3 OH −( aq)3 H 2 O (l) ∆ H =n ∆ H ( products)−n ∆ H (reactants) ∆ H = [ 3 (−286 ) ] – [3 (0 ) +3 ( −230) ]

kJ kJ =−168 mol mol

kJ kJ −( −158 ) mol mol ¿ ¿ % Error=¿

(−168)

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Jiries Meehan-Atrash

Caitlin Bettenay

Wednesday 25th January 2017

Discussion: Is your value for the ∆H of neutralization for phosphoric acid greater than or less than the accepted value. Think of some valid sources of error to account for your difference and explain how they would contribute to the direction of your error. The value found for change in enthalpy of the neutralization of phosphoric acid (-168kJ/mol) was greater than the accepted value (-158kJ/mol). Using these values, the percentage of error was calculated and found to be 6%. Possible sources of error could include measuring the exact volume of the solution. A graduated cylinder was used to measure the solution but a slight inaccuracy can cause a change in the mass of the solution. Also, it was assumed that the specific heat of the solution was the same as water, this assumption could lead to calculation errors. Furthermore, if any of the other assumptions made were found to be untrue, then this would lead to significant calculation errors and the experiment would no longer be valid.

The Enthalpy of Neutralization Worksheet: The r ei snotaf or mall a br e por tf ort hi sl a b.Compl e t et heabo vepa ge sa nds ubmi tt he m t oyourTA i nt hedr opbox .

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