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Yuri [45]
3 years ago
8

Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. ????????????H(????) → ????????+(?????

???) + ????H −(????????) + x1????????
Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of hydrogen chloride to form water and an aqueous solution of sodium chloride. ????????????H(????) + H +(????????) + ????????−(????????) → H2????(????) + ????????+(????????) + ????????−(????????) + x2????????
Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of hydrogen chloride to form water and an aqueous solution of sodium chloride. ????????+(????????) + ????H −(????????) + H +(????????) + ????????−(????????) → H2????(????) + ????????+(????????) + ????????−(????????) + x3????J
Procedure Reaction
1 a. In the glassware menu, take out a 50 mL graduated cylinder and a foam cup. From the tools menu, take out the scale. From the solutions stockroom, move the distilled water and solid NaOH onto the workbench.
b. Transfer 50.0 mL of water to the foam cup. To do this, drag the carboy of water onto the graduated cylinder. (Before you release the mouse button, the cursor will show a "plus sign" to indicate that it is the recipient). A transfer textbar will appear, enter "50.0" mL and click on pour. (You will notice that the graduated cylinder now reads 50.0 mL).
c. Weigh about 1 gram of solid sodium hydroxide pellets, NaOH(s), directly into the foam cup and record its mass to the nearest 0.01 gram. To do this, place the foam cup on the balance so it registers a mass, then click the "Tare" button. Drag the NaOH bottle onto the foam cup. (When you release the mouse, the bottle will be tipped to show that it is in the pour mode). Next, type "1.00" grams into the transfer bar and then click pour. Note that the balance now reads the mass of the transferred NaOH. You may now take the cup off of the scale.
d. Click on the graduated cylinder, record its temperature and then drag it onto the foam cup. (When you release the mouse, the graduated cylinder will be tipped to show that it is in pour mode.) Enter "50.0" mL in the transfer bar and then click pour. Record the highest temperature. e. Remove the foam cup and graduated cylinder from the workbench. (Right click on the item and select "remove.")
Reaction 2
a. Take the 0.5 M HCl from the strong acids cabinet and a fresh foam cup and a fresh 50 mL graduated cylinder from the glassware menu and place them on the workbench. The procedure for Reaction 2 is the same as for Reaction 1 except that 50.0 mL of 0.50 M hydrochloric acid solution is used in place of the water. After measuring 50.0 mL of the HCl solution into the graduated cylinder, proceed as before with steps b-e of the procedure for Reaction 1.
Reaction 3
a. Take out a 25 mL graduated cylinder, a fresh foam cup, the 1.0 M HCl and the 1.0 M NaOH. (If you are running out of room on the workbench, you may remove the previously used chemicals.) Use the graduated cylinder to measure and transfer 25.0 mL of 1.0 M HCl into the foam cup. Pour an equal volume of 1.0 M sodium hydroxide solution into a clean graduated cylinder.
b. Record the temperature of each solution to the nearest 0.1 oC. Pour the sodium hydroxide solution into the foam cup and record the highest temperature obtained during the reaction.
Data and Analysis
Reaction 1Reaction 2Reaction 3
Mass of solution* (g) 1.03g 1.03g
Initial temperature(°C) 25oC 25OC 25OC
Maximum temperature (°C) 30.3oC 37oC 31.7oC
Temperature change (∆T)
Heat energy q (kJ)
Moles of NaOH
Molar heat of reaction (-q/mol) also known as Enthalpy change,
DH (kJ/mol)
Chemistry
1 answer:
Dennis_Churaev [7]3 years ago
5 0
Hi, here is a basic summary of what we did in a lab; there were 3 reactions: The procedure: Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. NaOH(s)-> Na+(aq) + OH-(aq) ΔH1=-34.121kJ Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of HCl to form water and an aqueous solution of sodium chloride. NaOH(s) + H+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH2=-83.602kJ Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of HCl to form water an an aqueous solution of sodium chloride. H+(aq) + OH-(aq) + Na+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH3= -50.2kJ The ΔH values were calculated by dividing the heat gained by the number of moles (each reaction had 0.05moles of NaOH) The problem: Net ionic equations for reaction 2 & 3: 2: NaOH(s) + H+(aq) -> H2O + Na+(aq) 3: H+(aq) + OH-(aq) -> H2O i) In reaction 1, ΔH1 represents the heat evolved as solid NaOH dissolves. Look at the net ionic equations for reactions 2 and 3 and make similar statements as to what ΔH2 and ΔH3 represent. ii) Compare ΔH2 with (ΔH1 + ΔH3). Explain in sentences the similarity between these two values by using your answer to #5 above. Attempt at answering: i) Firstly, ΔH2 represents the heat evolved as the hydrogen ion displaces the sodium ion, creating a single displacement reaction. ΔH3 represents the heat evolved as the hydrogen and hydroxide ion form water via a neutralization reaction. ii) ΔH2 is equal to (or supposed to be, this is a source of error while calculating) (ΔH1 + ΔH3). The similarity between these two values is that .. (this is where I get confused!)

Source https://www.physicsforums.com/threads/calorimetry-help-chemistry.399653/
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                   x                     --------------------- 3.7 x 10²⁰ molecules

                   x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³

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