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OR # Form 4 Chemistry: Energy changes in chemical and physical processes lessons

Experiment to determine the heat of neutralization of hydrochloric acid by sodium hydroxide.

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Experiment: - To determine the heat of neutralization of hydrochloric acid by sodium hydroxide:
(i) Procedure:-
- A clean dry 250 cm^3 glass or plastic beaker is wrapped with a newspaper leaf.
- Exactly 50 cm^3 of 2.0M hydrochloric acid solution is transferred into the beaker.
- The temperature T1 of the acid solution is noted.
- Using another clean dry measuring cylinder, exactly 50cm^3 of 2.0M NaOH solution is measured and its steady temperature T_2 is noted.
- The contents of the beaker (acid), are carefully stirred with a thermometer while adding NaOH.
- The highest temperature T_4 attained by the resulting mixture is noted.
(iii). Results: - Temperature of the acid, T_1 ^0C
- Temperature of the hydroxide, T_2 ^0C;
- Average temperature of the two solutions; (T_1 + T_2)/2 = T_3^0C
-The highest temperature of the mixture; T4 ^0C;
-The temperature change, ΔT= (T4 –T3) ^0C;
Sample calculations: Given: Temperature of hydrochloric acid solution T1= 22.75^0C;
Temperature of sodium hydroxide solution T2 = 22.80^0C; Average temperature of acid and alkali Highest temperature of alkali and acid mixture, T4 =36.40^0C; Temperature change, ΔT= T4 – T3 = (36.40 – 22.78)^0C;
=13.62^0C; Assumption: the specific heat capacity of the solution= 4.2KJKg^-1K^-1.
- In the experiment, 50 cm^3 of 2M HCl are neutralized by 50 cm^3 of NaOH, thus; volume of the mixture= (50 + 50) cm^3 = 100 cm^3.
- Taking density of the resultant solution to be 1gcm^-3, then;-
Mass of solution M; =density x volume =1g/cm3 x 100cm^3= 100g.=0.1 kg i.e. (100/1000)
Thus heat evolved, Mass of solution x specific heat capacity x temperature change = MCΔT, = 0.1Kg x 4.2kj/kg/k x 13.62^0C =- 5.7 KJ.
- But 50 cm^3 of 2M HCl contains 2 x 50 moles, = 0.1 moles of H+ ions. 1000
- Similarity 50 cm^3 of 2M NaOH contains 2 x 50 = 0.1 moles of OH- ions.1000
- This implies that the two solutions neutralize each other completely.
Therefore;
Molar heat of neutralization (heat liberated when one mole of each reagent is used;- 0.1 Mole = 5.72KJ /1mole = 1 x 5.72 = -57.2 KJMol^-1
- Since heat is evolved, the reaction is exothermic, the molar heat of neutralization= -57.2 KJMol^-1.
Thus, enthalpy change; H^(+) (aq) + OH^(-) (aq) to H_2O(l); ΔH(neut) = - 57.2 KJMol^-1.
Note:
Thermochemical equation: refers to a chemical equation which shows the enthalpy change.
(c) Calculate;
(i) The amount of heat produced during the experiment. (Specific heat capacity of solution=4.2kjkg^-1k^-1, density of solution= 1gcm^- 3).
Solution:
Amount of heat = MCΔT Mass of solution = (145.1 – 45.1) = 100g
Temperature change; ΔT =(38.5)-(27.0 +23.0) = 38.5 –25 =13.5^0C;
Heat produced ΔH = 100g x4.2kjg^-1k^-1 x 13.5^0C=5670 joules=5.67KJ
(ii). Molar heat of neutralization for the reaction.
Solution:
Number of moles involved;taking only NaOH or HCl’ 1000 cm^3 = 2moles 50 cm^3= 2 x 50 = 0.1moles 1000
If 0.1 mole produced 5.67 kg; 1 mol = ?0.1=5.67kg/ 1mol= (5.67 x 1) = 56.7 KJMol^-1
(d). Explain why the molar heat of neutralization of NaOH and ethanoic acid of equal volume and molarity would be less than the value obtained in c (ii)
above.
Solution:
- Some of the heat produced during neutralization is used up by the weak acid to dissociate fully hence the lower value.
(e). Write down the Thermochemical equation for the reaction between NaOH and dilute hydrochloric acid above. Solution NaOH(aq) + HCl(aq) NaCl(aq) + H_2O(l); ΔH= -56.7KJMol^-1