The enthalpy change for a reaction between a strong acid and a strong alkali

Aim / Objective: 

To determine the enthalpy change for a reaction between a strong acid and a strong alkali.

Introduction:  Enthalpy is defined as the total energy in a system.  The change in energy ∆H can be positive in heat absorbing (endothermic reactions) or negative in heat releasing (exothermic reactions). This experiment focuses on one form of enthalpy change which is enthalpy of neutralization (∆Hn).  Enthalpy of Neutralization is the enthalpy change observed when one mole of water is formed when a base reacts with an acid in a thermodynamic system.

The literature standard enthalpy for a strong acid-base reaction is -57.1kJ/mol. For weak acids and bases the heat of neutralization is different as they are not fully dissociated and hence some heat will be absorbed.

Materials/ Apparatus:

1.0M HCl solution, 1.0M NaOH solution, 2 measuring cylinders, 2 Styrofoam cups, 2 beakers, 1 thermometer, 1 glass stirring rod.

Method / Procedure:

  1. Use a measuring cylinder to measure 50cm3 of sodium hydroxide (NaOH) and pour into a Styrofoam cup.
  2. Use a thermometer to measure the temperature of the NaOH. Record the reading.
  3. Use a measuring cylinder to measure 50cm3 of hydrochloric acid (HCl) and pour into a Styrofoam cup.
  4. Use a thermometer to measure the temperature of the HCl. Record the reading.
  5. Mix the contents of both cups in a beaker and stir the contents using the glass stirring rod.
  6. Take the temperature after stirring the mixture then record the reading.
  7. Repeat steps 1 to 6 TWO times and then tabulate the results.

Suggested Results:

Experiment number Initial temperature / °C Final temperature / °C
1 32 33.5 46
2 30 30.5 46
3 30 30.5 46

Discussion / Calculations:

Heat released = mcΔT

  1. Calculate mass

1000g = 1kg

1000g ÷1000

x = 1kg

50cm3 of acid and 50cm3 alkali = 100cm3 = 1kg

  1. Average temperature = (32+30+30+33.5+30.5+30.5) / 6

= 186.5/6

= 31.08°C

  1. Change in temperature (ΔT) = final temperature – initial temperature

= (46 – 31.08) °C

= 14.92°C

Change °C to K = 14.92 + 273 =287.92

  1. Specific heat capacity = 4.187JK-1kg-1
  1. Heat energy released = mcΔT

= 1kg x 4.187JK-1kg-1 x 287.92K

= 1205.52104J

  1. Enthalpy change = mcΔT / number of moles

50cm3 of acid & base contains (2/1000) x 50

=0.1 moles

0.1 moles H20 = 727.92J

1.0 moles H2O = 727.92J / 0.1 moles

= 7279.2J

Change from J to kJ

7279.2J / 1000 = 7.279kJ

Write the molecular and ionic equation for the reaction

NaOH(aq)   +  HCL(aq)                NaCl(aq) +  H2O(l)

H+(aq)   +    OH(aq)                 H2O(l)

Explain whether the reaction was exothermic or endothermic?

Heat was lost from the mixture to the environment can be said to be exothermic.

What was the use of the Styrofoam cups?

They were used to minimize heat loss of the mixture to the environment.

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