Answer:
determination 1
1) Temperature change for the reaction T = T final - T initial
= 34.25 - 21.27 [ average initial temp of acid and base = 21.27]
= 12.98 deg C
2) Amount of heat released = Total volume x density x sp.heat x T
= 101 mL x 1.04 g / mL x 3.89 J / g-deg C x 12.98 deg C
= 5304 J
3) mol of HCl = Vol of HCl added mL x 1 L / 1000 mL x Molarity of HCl mol / L [ as per eq.5 ]
= 50 mL x 1 L / 1000 mL x 2 mol/L
= 0.1 mole
similarly mol of NaOH = 51 mL x 1 L / 1000 mL x 2 mol / L
= 0.102 mol
4 ) mole of NaOH reacted = 0.1 mol [ since here HCl is the limiting reagent which has 0.1 mole only.]
5) Heat of neutralisation = - amount of energy released / number of moles of NaOH reacted
= - 5304 J / 0.10 mol
= - 53040 J / mol = - 5.3 x 104 J / mol
determination 2
1) Temperature change for the reaction T = T final - T initial
= 34.24 - 21.11 [ average initial temp of acid and base = 21.3]
= 13.13 deg C
2) Amount of heat released = Total volume x density x sp.heat x T
= 101 mL x 1.04 g / mL x 3.89 J / g-deg C x 13.13 deg C
= 5365 J
3) mol of HCl = Vol of HCl added mL x 1 L / 1000 mL x Molarity of HCl mol / L [ as per eq.5 ]
= 50 mL x 1 L / 1000 mL x 2 mol/L
= 0.1 mole
similarly mol of NaOH = 51 mL x 1 L / 1000 mL x 2 mol / L
= 0.102 mol
4 ) mole of NaOH reacted = 0.1 mol [ since here HCl is the limiting reagent which has 0.1 mole only.]
5) Heat of neutralisation = - amount of energy released / number of moles of NaOH reacted
= - 5365 J / 0.10 mol
= - 53650 J / mol = - 5.37 x 104 J / mol
Thus
the heat of neutralisation for determination 1 = 5.3 x 104 J / mol
for determination 2 = 5.37 x 104 J / mol
44. For each determination calculate the number of moles of HCl and of Naon that were...
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