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A reaction vessel of negligible heat capacity held 75 cm³ of 1.0 M HCl solution at a temperature of 13.0 °C - Leaving Cert Chemistry - Question 3 - 2013

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A reaction vessel of negligible heat capacity held 75 cm³ of 1.0 M HCl solution at a temperature of 13.0 °C. A thermometer was placed in the liquid in the reaction v... show full transcript

Worked Solution & Example Answer:A reaction vessel of negligible heat capacity held 75 cm³ of 1.0 M HCl solution at a temperature of 13.0 °C - Leaving Cert Chemistry - Question 3 - 2013

Step 1

Define heat of reaction.

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Answer

The heat of reaction is defined as the heat change that occurs when the number of moles of reactants in a balanced equation react completely. It is quantified as the heat change for the reaction according to the balanced equation.

Step 2

Suggest a suitable material for the reaction vessel to avoid heat loss to the surroundings.

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Polystyrene or other suitable insulating materials such as plastic or cardboard should be used as they minimize heat transfer to the environment.

Step 3

State (i) one advantage, (ii) one disadvantage, of the use of a burette instead of a graduated cylinder for measuring out the base and adding it to the reaction vessel.

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(i) One advantage of using a burette is that it allows for more precise delivery of the reactant, enabling more accurate measurements. (ii) One disadvantage is that a burette may be more difficult to read accurately compared to a graduated cylinder, particularly if the level of the liquid is very close to a graduation line.

Step 4

State two ways of ensuring that the rise in temperature was measured as accurately as possible.

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  1. Use a sensitive thermometer that can read to at least 0.2 °C, ensuring it is properly calibrated.
  2. Minimize external influences by covering the reaction vessel to prevent heat loss and ensuring the thermometer is not exposed to air drafts during measurements.

Step 5

How many moles of HCl were neutralized in the reaction with NaOH?

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To calculate the moles of HCl, we can use the formula:

n=CimesVn = C imes V

Where:

  • CC = concentration in mol/L = 1.0 M
  • VV = volume in L = 0.075 L (75 cm³ converted to liters)

Hence, n=1.0imes0.075=0.075extmoln = 1.0 imes 0.075 = 0.075 ext{ mol}

Step 6

Calculate the heat produced in the reaction vessel as a result of the reaction of the HCl with the NaOH.

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Using the formula for heat transfer:

q=mchetaq = mc heta

Where:

  • mm = mass of the solution = 75 g (since the density of solution is 1 g/cm³)
  • cc = specific heat capacity = 4.2 J/g°C
  • heta heta = change in temperature = 20.9 °C - 13.0 °C = 7.9 °C

Hence, q=75imes4.2imes7.9=2,478.75extJ=2.48extkJq = 75 imes 4.2 imes 7.9 = 2,478.75 ext{ J} = 2.48 ext{ kJ}

Step 7

Hence calculate the heat of reaction for the neutralisation reaction between hydrochloric acid and sodium hydroxide.

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To calculate the heat of reaction per mole of HCl:

= rac{2.48 ext{ kJ}}{0.075 ext{ mol}} = 33.07 ext{ kJ/mol}$$

Step 8

Identify the hazards associated with the use of hydrochloric acid and sodium hydroxide.

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Hydrochloric acid is corrosive and can cause burns to the skin and eyes. Sodium hydroxide is also corrosive and can cause severe damage to skin and eyes upon contact.

Step 9

Describe how the experiment should be carried out to label the two solutions. What experimental problem would have been encountered if 0.1 M NaOH and 0.1 M HCl solutions had been used instead of 1.0 M solutions?

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Answer

To label the two solutions, proper labeling should be done before the experiment includes the correct concentration, compound name, and hazard symbols. If 0.1 M solutions were used, the problem would be that the reaction would be slower and may not reach measurable heat changes, leading to inaccurate results.

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