SHOW ALL WORK!! Assume that the experiment is repeated but the volumes of NaOH a
ID: 989034 • Letter: S
Question
SHOW ALL WORK!!
Assume that the experiment is repeated but the volumes of NaOH and HCl are doubled. (100.0 mL of 1.0 M HCl and 100.0 mL of 1.0 M NaOH) The NaOH is reacted with HCl. How would the calculated values of each of the following compare to the values that were obtained when you completed the experiment?
Orignal values:
average temp of HCl: 22.2
Average temp of NaOH: 23
mixed at 4th min
5th-20th min: 25.6, 25.8, 26, 26.3, 26.5, 26.8, 27.1, 27.4, 27.7, 28, 28.3, 28.5, 28.8, 29, 29.3, 29.6
I think.......
heated gained: 2856
Heat of reactuion: -2856
Enthalpy of neutralization: 57120
Explanation / Answer
Total volume of mixture= 100+100ml =200ml
Demsity of solution =1 g/cc ( assumed), mass of solution =200*1 =200 gm tihs contains 100 gms of HCl and 100gm of NaOH
heat gained due to the reaction= 100*4.188(29.6-22.3) ( this is Contribution due to HCl)+ 100*4.18*(29.6-23)( contribution of NaOH)=5810.2 Joule
moles of HCl = 1*100/1000 =0.1
Heat of neutralization= 5810.2/0.1= 58102 Joules
The reaction is HCl+ NaOH---- > NaCl+ H2O
1 mole of HCl and 1 mole of NaOH gives rise to 1 mole of NaCl and 1 mole of water.
2 moles of reactants are giving 1 mole of NaCl.
Heat gainaed= 58102/2= 2905.1 Joules
Heat of reaction= -2905.1 Joules ( since during the course of reaction heat is liberated and heat gained = heat of reaction)
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