The following results were obtained in an experiment to determine the heat of neutralization of 50cm3 2M hydrochloric acid and 50cm3 2M sodium hydroxide. Mass of...

      

The following results were obtained in an experiment to determine the heat of neutralization of 50cm3 2M hydrochloric acid and 50cm3 2M sodium hydroxide.

Mass of plastic cup = 45.1g
Initial temperature of acid = 27.00C
Initial temperature of alkali = 23.00C
Mass of plastic cup + HCl + NaOH = 145.1g
Temperature of the mixture of acid and alkali = 38.50C

(a) Define heat of neutralization

(b) Write an ionic equation for the neutralization of hydrochloric acid and sodium hydroxide.
(i) The amount of heat produced during the experiment
(Specific heat capacity of solution = 4.2jg-1k-1, density of solution = 1gcm3)

(ii) Molar heat of neutralization for the reaction.

(c) 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.

(d) Write down the thermochemical equation for reaction between NaOH and dilute hydrochloric acid
above

(e) Draw an energy level diagram for the neutralization reaction in (d) above.

  

Answers


Maurice
(a) Heat of Neutralization is the enthalpy change when 1 mole of hydrogen ions from an acid react with
1 mole of ions from an alkali/base to form one mole of water.

(b) Q7942019905.png
Q7942019908.png
Q7942019910.png

(c) Some of the energy is used to ionize the weak Ethanoic acid.

(d) NaOH(aq) + HCl(aq)----->NaCl(aq) + H2O(1)

(e) Q7942019918.png
maurice.mutuku answered the question on September 4, 2019 at 06:20


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