IJSO – Training Material Phase 3 – Practical Training

Experiment 1

Investigating the Effects of Concentration and Temperature on the Rate of a Chemical Reaction

Objective:To investigate the concentration and temperature effects on the rate of a chemical reaction

Introduction

You have learnt in the classroom session that the rate of a chemical reaction depends on the concentration of the reactants as well as the reaction temperature. We are going to investigate this chemical phenomenon. To facilitate our investigation, we need to choose a chemical reaction that exhibits some easily observable physical changes so that we can monitor the progress of the reaction conveniently. In this regard we will study the reaction between sodium thiosulphate and diluted hydrochloric acid.

S2O32-(aq) + 2 H+(aq)  SO2(g) + S(s) + H2O(l)

Sodium thiosulphate reacts with diluted acid to give sulphur dioxide, sulphur and water. Both sodium thiosulphate and diluted hydrochloric acid are colorless solution. Sulphur dioxide is a very soluble gas and dissolves completely in the aqueous solution. The sulphur formed, however, is not soluble and will exist in the mixture as white or yellow precipitate (or colloidal). It makes the reaction mixture becomes opaque as the reaction occurs. Therefore, we can study the reaction rate by monitoring the opaqueness of the reaction. This can be easily done by measuring the time taken for forming a certain amount of precipitate.

In the experiment, you will carry out the reaction by mixing the two reactants into a small beaker, and put the beaker on top of a piece of white paper which has a mark on it. Before the reaction starts you can see the mark clearly from the top of the beaker through the solution. However, as the reaction proceeds, sulphur precipitate forms and makes the solution become more opaque, and so eventually the mark is completely masked. The time taken for the mark to become totally disappeared indicates how fast the reaction has occurred.

Chemicals:0.15M sodium thiosulphate solution

1M hydrochloric acid

Apparatus:100 mL beakers

100 mL measuring cylinder

Thermometer (alcohol, up to 110oC)

Stopwatch

Part A: To study the concentration effect on the rate of a chemical reaction.

In this part of the experiment you will study the concentrate effect on the rate of the above reaction by varying the concentration of the sodium thiosulphate solution.

Procedure:

  1. Take a piece of white paper and make a clear mark on it.
  2. Put a dry clean beaker on top of the mark.
  3. Measure 60 mL of 0.15 M sodium thiosulphate solution and put it into the beaker.
  4. Quickly add 10 mL of 1 M HCl into the beaker in a single portion and start the stopwatch immediately.
  5. Stir the mixture gently with a glass rod for a few seconds.
  6. Look at the mark from vertically above the beaker through its content, and record the time taken for the mark to become totally disappeared.
  7. Repeat steps 1-6 by substituting the 60 mL 0.15 M sodium thiosulphate solution with:

Solution / Vol. of 0.15M Na2S2O3 solution / Vol. of water / Total volume
A / 45 mL / 15 mL / 60 mL
B / 30 mL / 30 mL / 60 mL
C / 15 mL / 45 mL / 60 mL

Data

  1. Time for the disappearance of the cross

Volume of Na2S2O3(aq) (mL) / Volume of water
(mL) / Volume of HCl(aq)
(mL) / Initial concentration of Na2S2O3(aq) in the reaction mixture/M / Time for the disappearance of the cross/s
Trial 1 / Trial 2 / Average
60 / 0 / 10 / 0.1500M / 16.6 / 17.4 / 17.00
45 / 15 / 10 / 0.1125M / 24.7 / 25.0 / 24.85
30 / 30 / 10 / 0.0750M / 36.6 / 37.1 / 36.85
15 / 45 / 10 / 0.0375M / 83.4 / 85.0 / 84.20

Data Analysis

  1. Calculate the initial concentration of sodium thiosulphate in the reaction mixtures.

Run 1

Initial concentration of Na2S2O3

= M

= 0.1500 M

Run 2

Initial concentration of Na2S2O3

= M

= 0.1125 M

Run 3

Initial concentration of Na2S2O3

= M

= 0.0750 M

Run 4

Initial concentration of Na2S2O3

= M

= 0.0375 M

  1. Plot the time required for the disappearance of the cross against the initial concentration of sodium thiosulphate on a graph paper. Label your graph properly.

  1. How is the rate of reaction affected by the concentration of the reactant?

The rate of reaction increases when the concentration of sodium thiosulphate increases.

Part B: To study the temperature effect on the rate of a chemical reaction.

In this part of the experiment you will study the temperature effect on the rate of the about reaction. You will carry out the reaction at different temperature and compare the results.

Temperature:0oC (ice-water bath)

Room temperature (refer to the data taken in Part A)
50oC (by warming the Na2S2O3 solution with a 50oC water bath)

Procedure

Reaction at 0oC

  1. Measure 60 mL of 0.15M sodium thiosulphate solution and put it into a clean, dry 100 mL beaker.
  2. Cool the solution in an ice-water bath for 15 minutes.
  3. Measure the temperature of the solution.
  4. Carry out the reaction of the cooled Na2S2O3 solution with 1M HCl by following the steps 1-6 described in part A.
  5. Measure 60 mL of 0.15M sodium thiosulphate solution and put it into another dry, clean 100 mL beaker.
  6. Warm the solution in a 50oC water bath for 15 minutes.
  7. Measure the temperature of the warmed solution.
  8. Carry out the reaction of the warmed Na2S2O3 solution with 1M HCl by following the steps 1-6 described in part A.

Data

Time for the disappearance of the cross

Initial concentration of Na2S2O3(aq) in the reaction moisture/M / Temperature of the reaction mixture/oC / Time for the disappearance of the cross/s
Trial 1 / Trial 2 / Average
0.15M / 51.5 / 5.9 / 5.7 / 5.80
0.15M / 23.8 / 16.6 / 17.4 / 17.00
0.15M / 7.8 / 46.0 / 45.3 / 45.65

Data Analysis

  1. From the data in the above table, how is the rate of reaction affected by temperature?

The rate of reaction increases when the temperature of reaction increases.

Experiment 2

Investigating the effect of surface area (particle size of a solid reactant) on the rate of a chemical reaction.

Objective:To investigate the effect of surface area on the rate of a chemical reaction

Introduction

If a chemical reaction involves one or more solid reactants, the particle sizes of the solid reactants will affect the rate of the reaction. It is because surface area increases as the particles become smaller. Increasing the reactants’ surface area allows the reactants approach to each other more frequently, and so the reaction rates are often enhanced. A daily example is the burning of charcoal. Powdered charcoal burns more fiercely than large lumps of charcoal. It is because powdered charcoal has a much larger surface for reacting with oxygen.

Zinc metal reacts with diluted hydrochloric acid to give zinc chloride and hydrogen gas. The rate of hydrogen gas formation allows us to observe the reaction rate conveniently. In this experiment we will compare the reaction rates of diluted hydrochloric acid with zinc powder and zinc granules.

Chemicals:1M hydrochloric acid

Zinc powder

Zinc granules

Apparatus:test tubes, spatula

Procedure:

  1. Measure about 5 mL of 1M hydrochloric acid into a test tube and add some zinc granules. Record the observation.
  2. Measure about 5 mL of 1M hydrochloric acid into another clean test tube and add zinc powder. The amount of zinc powder added should be approximately equal to the amount of zinc granules added in step 1. Record the observation.
  3. Compare the rate of these two reactions.

Data Analysis

  1. Which of the two reactions occurs faster? Did zinc granules or zinc powder give a faster reaction rate?

The reaction with zinc powder occurs faster, so zinc powder give a faster reaction rate.

  1. How does the particle size of solid reactant affect the reaction rate?

When the particle size of solid reactant smaller, its surface area increases, and allows reactants approach to each other more frequently. Thus, the reaction rate will be enhanced.

Experiment 3

Determination of the Rate Equation for a Chemical Reaction

Objective:To determine the rate equation for the reaction of iodine and propanone in acidic medium

Introduction

From experiment 1 you should have realized that the rate of a chemical equation depends on its reactants’ concentration. The relationship between the rate of a chemical reaction and its reactants’ concentration can be expressed into a rate equation. For example, propanone reacts with iodine in acidic medium as shown below:

CH3COOCH3(aq) + I2(aq) CH3COCH2I(aq) + H+(aq) + I-(aq)

The rate of the reaction may depend on the concentration of propanone (CH3COCH3), iodine (I2) and acid (H+). The rate equation of the reaction can be expressed in the form of:

Rate = k[CH3COCH3]a[I2]b[H+]c

where [CH3COCH3], [I2] and [H+] are the concentration of propanone, iodine and acid, respectively. k is the rate constant which is a quantity that depends on reaction temperature.a, b and c are the orders of reaction with respect to propanone, iodine and acid, respectively. Reaction orders (a, b and c) are quantities that must be determined experimentally and cannot be deduced from the chemical equation. The reaction order with respect to a reactant is not necessarily equal to the reactant’s coefficient present in the balanced chemical equation.

In this experiment you are going to determine the values of k, a, b, and c for the reaction of propanone with iodine. You will do so by running the reaction several times and varying the concentration of the reactants one at a time.

Chemicals:1M hydrochloric acid

1 M propanone

0.002M iodine solution (prepared by dissolving 0.5 g of I2 and 3.3 g of KI in 1 L of deionized water)

Apparatus:measuring cylinders, conical flasks

Procedure:

  1. Use measuring cylinders to measure the appropriate amounts of hydrochloric acid, propanone solution and deionized water into dry conical flasks, according to the table shown below.
  2. Add appropriate amounts of iodine solution into the conical flasks. Start the stop watch.
  3. Swirl the flasks gently.
  4. Measure the time taken for the colour of iodine disappears completely.

Data

Run 1 / Run 2 / Run 3 / Run 4
Volume of 1 M HCl(aq)(mL) / 20 / 10 / 20 / 20
Volume of 1 M CH3COCH3(aq)(mL) / 8 / 8 / 4 / 8
Volume of deionised water (mL) / 0 / 10 / 4 / 2
Volume of 0.002M I2(aq)(mL) / 4 / 4 / 4 / 2
 [I2] in the time taken (mol dm-3) / 2.5 10-4 / 2.5 10-4 / 2.5 10-4 / 1.2510-4
Time for the disappearance of the colour (s) / Trial 1 / 52.5 / 102.8 / 109.0 / 40.0
Trial 2 / 49.5 / 102.7 / 104.5 / 38.5
Average / 51.00 / 102.75 / 106.75 / 39.25
Initial rate of reaction (mol dm-3 s-1) / 4.9010-4 / 2.43 10-4 / 2.3410-4 / 3.1810-4

Data Analysis

  1. Calculate the initial concentration of I2 in the reaction mixture and hence the change in iodine concentration in the time taken.

Run 1

Initial concentration of I2

= M

= 2.510-4M

Run 2

Initial concentration of I2

= M

= 2.510-4M

Run 3

Initial concentration of I2

= M

= 2.510-4M

Run 4

Initial concentration of I2

= M

= 1.2510-4M

  1. Calculate the rate of consumption of I2, this is the initial rate of reaction in terms of I2.

Run 1

Initial rate of reaction in term of I2

=

= 4.9010-6 mol dm-3 s-1

Run 2

Initial rate of reaction in term of I2

=

= 2.4310-6 mol dm-3 s-1

Run 3

Initial rate of reaction in term of I2

=

= 2.3410-6 mol dm-3 s-1

Run 4

Initial rate of reaction in term of I2

=

= 3.1810-6 mol dm-3 s-1

  1. Determine the order of reaction with respect to each reactant.

Rate = k[CH3COCH3]a[I2]b[H+]c

Compare Run 1 and Run 2,

[I2][CH3COCH3] are the same in these runs by using the same volume, and they have the same k value.

 The rate equation can be writted as Rate = k[H+]a

2.16= [2]c

log 2.16 = c log 2

c = log 2.16 / log 2

= 1.11

~ 1

 The order of H+ is 1

Compare Run 1 and Run 3,

[I2][H+] are the same in these runs by using the same volume, and they have the same k value.

 The rate equation can be writted as Rate = k[CH3COCH3]a

2.09 = [2]a

log 2.09 = a log 2

a = log 2.09 / log 2

= 1.06

~ 1

 The order of CH3COCH3 is 1

Compare Run 1 and Run 4,

[CH3COCH3][H+] are the same in these runs by using the same volume, and they have the same k value.

 The rate equation can be writted as Rate = k[I2]b

1.54 = [2]a

log 1.54 = c log 2

b = log 1.54 / log 2

= 0.62

1

 The order of I2 is 0

  1. Estimate the rate constant (with unit) for the reactionusing the data in Run 1.

Rate = k[CH3COCH3]1[I2]0[H+]1

Rate = k[CH3COCH3][ H+]

4.9010-6= k [(10.008/0.032][ (10.020/0.032]

k = 3.13610-5 mol-1 dm-3 s-1

  1. Write the rate equation for the reaction.

Rate = 3.13610-5 [CH3COCH3]1 [H+]1

1