(a) write a balanced chemical equation for the reaction that takes place in a st
ID: 813063 • Letter: #
Question
(a) write a balanced chemical equation for the reaction that takes place in a styrofoam cup.
(b) Is any NaOH or H2SO4 left in the cup when the reaction is over?
(c) Calculate the enthalpy change per mole of H2SO4 in the reaction
_________________ kJ/mol
A 100.0 mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 22.1 degree C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 32.7 degree C. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g x degree C), and that no heat is lost to the surroundings. (a) write a balanced chemical equation for the reaction that takes place in a styrofoam cup. (b) Is any NaOH or H2SO4 left in the cup when the reaction is over? (c) Calculate the enthalpy change per mole of H2SO4 in the reaction _________________ kJ/molExplanation / Answer
A
2NaOH + H2SO4 ---> Na2SO4 + 2H2O
100 mL* 1 M = 100 mmol NaOH
50 mL * 1 M = 50 mmole H2SO4
so the actual mole ratio matches the stoichiometric ratio of 2 to 1
B
. Moles NaOH: 0.100 L x 1.00M = 0.100 moles base
Moles H2SO4: 0.050 L x 1.00M = 0.050 moles acid
Mole ratio base to acid 2:1,
0.050 moles acid needs x2 base, there is enough
for both to completely react into products, so none left
C
the total solution gains the heat given up by the reaction
so
Q = m c delta(T) = 150.0 g (4.18 J/(g C) (32.7 - 22.1) = 6646.2 J
then
6646.2 J / 50E-3 mol = 132924 J/mol or 132.924 kJ/mol
to 3 s.f. it's -133. kJ/mol
negative because the reaction is exothermic
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