When 50 0 ml. of 1.00 M H_2SO_4(aq), at 26.1 degree C was added to 50.0 mL of 1.

Question

When 50 0 ml. of 1.00 M H_2SO_4(aq), at 26.1 degree C was added to 50.0 mL of 1.00 M NaOH also at 26.1 degree C, the temperature rose to 32.6 degree C. Assume the resulting solution had a total volume of 100.0 mL with the same density and specific heat as water. Calculate Delta H for the reaction described by this equation. H_2SO_4(aq) + 2 NaOH(aq) rightarrow Na_2SO_4(aq) + 2 H_2O(l) To do this calculation, work through the following questions and steps (a-f) Which reactant limits? How many moles of it are present? How much energy is required to heat 100.0 g solution from 26.1 degree C to 32.6 degree C? How much energy is released by the reaction per one mole of the limiting reactant? How many moles of that reactant are in the given equation? What is Delta H for the reaction as written above in kJ (not kJ/mol)?

Explanation / Answer

mmol of acid = MV = 50*1 = 50 mmol H2SO4

mmol of base = MV = 50*1 = 50 mmol NaOH

we need 2:1 so

NAOH will be limiting, since 25 mmol of H2SO4 will be left

b)

50 mmol or 50*10^-3 mol are present of NaOH

c)

assume this is only water so

Q = m*Cp*(Tf-Ti)

Q = (50+50)*4.184*(32.6-26.1) = 2719.6 J

d)

1 mol of NaOH

this was n = 50 mmol of NaOH = 50*!0^-3

then

E = -Q/n = -(2719.6)/(50*10^-3) = -54392J/mol = 54.39 kJ/mol

e)

50 mmol of NaOH or 50*10^-3 mol

f)

then we would not have to divide by moles

moles = 50mmol or 50*10^-3

Q = 2719.6 J or 2.71 kJ

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