A 90.0-mL sample of 1.00 M NaOH is mixed with 45.0 mL of 1.00 M H_2SO_4 in a lar

Question

A 90.0-mL sample of 1.00 M NaOH is mixed with 45.0 mL of 1.00 M H_2SO_4 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 23.7degreeC. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 30.5degreeC. 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 degreeC), and that no heat is lost to the surroundings. Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. Is any NaOH or H_2SO_4 left in the Styrofoam cup when the reaction is over? Include phases in the balanced chemical equation. Type an open parenthesis "(" to add a phase. Phases should not be subscripted. Use the left and right arrow keys to move the cursor out of a superscript or subscript in the module. Calculate the enthalpy change per mole of H_2SO_4 in the reaction.

Explanation / Answer

a)

the balanced reaction is given by

H2S04 (aq) + 2 NaOH (aq) ---> Na2S04 (aq) + 2H20 (l)

b)

we know that

moles = molarity x volume (L)

so

moles of NaOH taken = 1 x 90 x 10-3 = 90 x 10-3

moles of H2S04 taken = 1 x 45 x 10-3 = 45 x 10-3

now

consider the reaction

H2S04 (aq) + 2 NaOH (aq) ---> Na2S04 (aq) + 2H20 (l)

we can see that

moles of NaOH reacted = 2 x moles of H2So4 taken

so

moles of NaOH reacted = 2 x 45 x 10-3 = 90 x 10-3

so

all the NaOH and H2S04 is reacted

so

No , H2S04 or NaOH is left

c)

total volume = 90 + 45 = 135

now

mass = density x volume

so

mass of solution = 135 x 1 = 135 g

now

heat = m x s x dT

so

Q = 135 x 4.18 x ( 30.5 - 23.7)

Q = 3837.24 J

now

Qrxn = - Q = -3837.24

now

enthalpy change = -Q / moles of H2S04

enthalpy change = -3837.24 / 45 x 10-3

enthalpy change = -85.272 x 1000

enthaly change = -85.272 kJ /mol

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