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(b) Calculate how much heat (in joules) transferred between the system and the s

ID: 496661 • Letter: #

Question

(b) Calculate how much heat (in joules) transferred between the system and the surroundings in this experiment.

(c) Calculate the molar heat of the neutralization reaction (in kJ/mol) (d)Is the neutralization reaction endo or exothermic? Can you suggest one reason why this can be true?

(d)Is the neutralization reaction endo or exothermic? Can you suggest one reason why this can be true?

5. In the second experiment, the student mixed in a coffer cup calorimeter 35.0 mL of 0.70M H2SO4 solution and 65.0 mL of 0.50M NaOH solution. Do you think that the change in temperature in the second experiment was similar to the one recorded in the first experiment? Why or why not?

4. Below is the graph that the student attached to her report. NaOH HCI 26 25.5 24.5 p 24 23.5 23 22.5 22 21.5 10 Time (min) Let's think how we can use it to calculate the heat of neutralization reaction. Notice: In this experiment, the student measured the change in temperature not for the system itself but for the surroundings (solution). The solution is diluted and we can safely consider its specific heat and its density to be the same as for pure water: c 4.18 J/g-oCand d 1.00 g/cm3 (a) What is the change in temperature recorded in this experiment? AT Did the system loose or gain energy in this experiment? Explain

Explanation / Answer

(a)

Over the 10 minutes duration, the temperature changes from 25.375 0C to 24.375 0C

Thus the temperature decrease equals = 25.375 - 24.375 = 1 0C

Since the temperature of the surroundings decreased, so the surroundings lost energy, which means that the system gained energy.

(b)

Assuming the volume of the solution as 'x' mL

So, mass of solution = x*1 = 'x' g

So, heat transferred = x*4.18*1 = 4.18*x Joules

(c)

You have not given complete data to solve this question.

(5)

Moles of acid = 0.035*0.7 = 0.0245

Moles of base = 0.065*0.50 = 0.0325

Here, 1 mole acid reacts with 2 moles base. So base is in limiting amount

Thus, 0.0325 moles of water are produced.

The temperature change would be the same if the same number of moles of water were produced in the first experiment as well.