1. Give chemical (not ionic) equations for the reactions that will occur during

Question

1. Give chemical (not ionic) equations for the reactions that will occur during this experiment. OHT 3 H20 2. What are the sources of heat released and absorbed in this experiment? If water is one of the products, how will it affect your calculation? Heat absor is Werter. a of neat releases a commom c hon af auch and 3. When solutions of two reactants were mixed in a coffee-cup calorimeter, the following temperatures were recorded as a function of time. The initial temperature, t, was 24.3°C. a. Plot the data on graph paper provided. Th the final temperature, by extrapolating the data to the time of mixing (time-0 s) with a straight line. time (s) T time (s) T (eCl 40.5 150 30 39.8 90 40.7 210 40.3 120 240 40.2 A0.6

Explanation / Answer

Prelab

1. molecular equations,

a) HCl + NaOH --> NaCl + H2O

b) CH3COOH + NaOH --> CH3COONa + H2O

2. Heat is released by the acid-base reaction and absorbed by water.

3. Reaction is exothermic, heat released.

4. Safety : chemicals are corrosive, so proper clothing and running reactions in well ventillated hood is suggested.

Enthalpy calculations

NaOH + HCl

Trial 1 : dH = 100 x 4.184 x 14.5/2 x 0.05 = 60.67 kJ/mol

Trial 2 : dH = 100 x 4.184 x 13.2/2 x 0.05 = 55.23 kJ/mol

Average = 57.95 kJ/mol

CH3COOH + NaOH

Trial 1 : dH = 100 x 4.184 x 13.2/2 x 0.05 = 55.23 kJ/mol

Trial 2 : dH = 100 x 4.184 x 12.8/2 x 0.05 = 53.55 kJ/mol

Average = 54.39 kJ/mol

Postlab

a. enthalpy

dH = -57.95 + 54.39 = 3.56 kJ/mol

b. % error = (3.56 - 1.7) x 100/3.56 = 52.25%

3. Specific heat of calorimeter

= [50 x 4.184 x (98 - 58) - 50 x 4.184 x (58 - 20)/(58 - 20)

= 11.01 J/oC

So the value we used 1 x 10^1 J/oC is correct

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