Question

100mL of 3M NaOH is poured into 100mL of 3M HF. The highest temperature reached is 48.71 degrees Celcius.

3M NaOH inital temp= 25 degrees Celcius

3M HF inital temp= 25 degrees Celcius

a. How many moles of Hf reacted? How many moles of NaOH?)

b. Assuming the specific heat of the mixed solution is 4.18 J/(g*C), and that the density of the mixed solution is 1.00 g/mL, use q = CDT to determine the heat added to the solution.

c. Heat evolved by the reaction is the negative of heat added to the solution. Find heat evolved per mol of HF that reacts.

d. Compare the %deviation of the result of c. to the result of Part 1 for this reaction.

Does the highest temperature reached (48.71) show accurate heats of reaction?

Answer 1

Given data

3 M NaOH = 3 moles NaOH/ L of solution

3 M HF = 3 moles HF/L of solution

100 ml of NaOH and 100 ml HF reacts

A ) Number of moles of NaOH in 100 ml =

3(100) /1000 = 0.3 moles

Moles NaOH reacted = 0.3 moles

Moles of HF in 100 ml = 3(100) /1000= 0.3

Moles of HF reacted = 0.3 moles

B) Cp(solution) = 4.18 J/g°C

Density of mixed solution = 1 g/ml

Initial temperature of mixture = 25°C

Final temperature of mixture = 48.71°C

Amount of heat = Q = mCp∆T

m = density×volume = 1(100+100)= 200 g

Q = 200(4.18) (48.71-25) = +19821.56 J

C) Assuming no heat losses heat added to system is equal to heat evolved by system = +19821.56J

Heat evolved = -9910. 78 for 0.3 mole of HF that reacts

For 1 mole of HF reacted ,heat evolved = (-19821.56(1) /0.3 = -66071.866 J/ mol HF reacted

D)

Volume of HF to be added to make 1 mole react = 1000/3 = 333.33 ml

Volume of NaOH added = 333.33 ml

Total volume = 333.33+333.33 = 666.66 ml

Density = 1g/ml

Mass = 1(666.66) = 666.66 g

Heat added = 66071.866 J

Cp = 4.18 J/g°C

Using heat balance

66071.866= 666.66(4.18) (T-25)

T = 48.71°C

From part A

T = 48.7 °C

There is no percent deviation and both maximum temperature remain same

Hence heats of reaction is accurate

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