A 62.0 mL sample of 1.0 M NaOH is mixed with 55.0 mL of 1.0 M H_2SO_4 in a large
ID: 497123 • Letter: A
Question
A 62.0 mL sample of 1.0 M NaOH is mixed with 55.0 mL of 1.0 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 20.5 degree C. After adding the NaOH solution to the coffee cup, the mixed solutions are stirred until reaction is complete. 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 middot degree C), and that no heat is lost to the surroundings, The Delta H for the neutralization of NaOH with H_2SO_4 is -114 kJ/mol H_2SO_4. What is the maximum measured temperature in the Styrofoam cup? Number 34.95 degree CExplanation / Answer
[NaOH] = 62 mL x 1.0 M = 62 mmol
[H2SO4] = 55 mL x 1.0 M = 55 mmol
2 NaOH + H2SO4 --------> Na2SO4 + 2 H2O
62.0 mmol of NaOH would react completely with 62.0 x (1/2) = 31 mmol of H2SO4, but there is 55 mmol H2SO4 present . Hence, H2SO4 is in excess and NaOH is the limiting reactant.
Excess H2SO4 = 55 mmol – 31 mmol = 24 mmol = 0.024 mol
Given that heat of neutralization = 114 kJ/mol H2SO4 = 114000 J/mol H2SO4
Then,
Heat produced in the neutralization reaction = (0.024 mol H2SO4) x (114000 J/mol H2SO4)
= 2736 J
This heat is absorbed by the water.
We know that Heat Q = msdT where m= mass s = specific heat dT = change in temperature
dT = Q/ms = Q / [ volume x density x s ]
Then,
Change in Temperature = 2736 J / [ (62.0 mL + 55.0 mL) x (1.00 g/mL)] x [4.18 J/(g·°C)]
= 5.6°C
Therefore,
Maxumum measured tempearature = Initial temperature + chnage in temperature
= 20.5 °C + 5.6 °C
= 26.1 °C
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