EduNinja

IB Chemistry SL2.2 Rate of chemical changeQuestion Bank

Question 1

[Maximum number: 10]

Reaction kinetics can be investigated using the iodine clock reaction. The equations for two reactions that occur are given below.

 Reaction A:H2O2(aq)+2I(aq)+2H+(aq)I2(aq)+2H2O(l) Reaction B:I2(aq)+2 S2O32(aq)2I(aq)+S4O62(aq)\begin{array}{ll} \text { Reaction } \mathrm{A}: & \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \text { Reaction } \mathrm{B}: & \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(\mathrm{aq}) \rightarrow 2 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}(\mathrm{aq}) \end{array}

Reaction B is much faster than reaction A , so the iodine, I2\mathrm{I}_{2}, formed in reaction A immediately reacts with thiosulfate ions, S2O32\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}, in reaction B , before it can react with starch to form the familiar blue-black, starch-iodine complex.

In one experiment the reaction mixture contained:
5.0±0.1 cm35.0 \pm 0.1 \mathrm{~cm}^{3} of 2.00 moldm32.00 \mathrm{~mol} \mathrm{dm}^{-3} hydrogen peroxide (H2O2)\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)5.0±0.1 cm35.0 \pm 0.1 \mathrm{~cm}^{3} of 1 % aqueous starch
20.0±0.1 cm320.0 \pm 0.1 \mathrm{~cm}^{3} of 1.00 moldm31.00 \mathrm{~mol} \mathrm{dm}^{-3} sulfuric acid (H2SO4)\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)20.0±0.1 cm320.0 \pm 0.1 \mathrm{~cm}^{3} of 0.0100 moldm30.0100 \mathrm{~mol} \mathrm{dm}{ }^{-3} sodium thiosulfate (Na2 S2O3)\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\right)50.0±0.1 cm350.0 \pm 0.1 \mathrm{~cm}^{3} of water with 0.0200±0.0001 g0.0200 \pm 0.0001 \mathrm{~g} of potassium iodide (KI) dissolved in it.
After 45 seconds this mixture suddenly changed from colourless to blue-black.

Question 1(h)

(a)

The colour change occurs when 1.00×104 mol1.00 \times 10^{-4} \mathrm{~mol} of iodine has been formed. Use the total volume of the solution and the time taken, to calculate the rate of the reaction, including appropriate units.

[ 4 ]

Question 1(i)

(b)

In a second experiment, the concentration of the hydrogen peroxide was decreased to 1.00moldm31.00 \mathrm{moldm}^{-3} while all other concentrations and volumes remained unchanged. The colour change now occurred after 100 seconds. Explain why the reaction in this experiment is slower than in the original experiment.

[ 2 ]

Question 1(j)

(c)

In a third experiment, 0.100 g of a black powder was also added while all other concentrations and volumes remained unchanged. The time taken for the solution to change colour was now 20 seconds. Outline why you think the colour change occurred more rapidly and how you could confirm your hypothesis.

[ 2 ]

Question 1(k)

(d)

Explain why increasing the temperature also decreases the time required for the colour to change.

[ 2 ]

Question 1

[Maximum number: 5]

The rate of the acid-catalysed iodination of propanone can be followed by measuring how the concentration of iodine changes with time.

I2(aq)+CH3COCH3(aq)CH3COCH2I(aq)+H+(aq)+I(aq)\mathrm{I}_{2}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq}) \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq})

Question 1(a)

Question 1(a)(i)

(a)
(i)

Suggest how the change of iodine concentration could be followed.

[ 1 ]

Question 1(a)(ii)

(ii)

A student produced these results with [H+]=0.15 moldm3\left[\mathrm{H}^{+}\right]=0.15 \mathrm{~mol} \mathrm{dm}^{-3}. Propanone and acid were in excess and iodine was the limiting reagent.

Determine the relative rate of reaction when [H+]=0.15 moldm3\left[\mathrm{H}^{+}\right]=0.15 \mathrm{~mol} \mathrm{dm}^{-3}.

Question image
[ 2 ]

Question 1(b)

(b)

The student then carried out the experiment at other acid concentrations with all other conditions remaining unchanged.

Table

State and explain the relationship between the rate of reaction and the concentration of acid.

[ 2 ]

Question 1

[Maximum number: 2]

Limestone can be converted into a variety of useful commercial products through the lime cycle. Limestone contains high percentages of calcium carbonate, CaCO3\mathrm{CaCO}_{3}.

Question image

Question 1(c)

Question 1(c)(i)

(a)
(i)

The potential energy profile for a reaction is shown. Sketch a dotted line labelled "Catalysed" to indicate the effect of a catalyst.

Question image
[ 1 ]

Question 1(c)(ii)

(ii)

Outline why a catalyst has such an effect.

[ 1 ]

Question 1

[Maximum number: 2]

Pasteurization is used to eliminate pathogenic bacteria. The concentration of vitamin C was monitored over a period of time in pasteurized and unpasteurized orange juice.

Question image

Question 1(a)

Question 1(a)(iii)

(a)
(i)

Calculate the average rate of decrease of vitamin C concentration for pasteurized juice, in μgcm3\mu \mathrm{g} \mathrm{cm}^{-3} day 1^{-1}, for the first 56 days.

[ 1 ]

Question 1(a)(iv)

(ii)

Deduce, referring to the graph, whether pasteurization affects the rate of change of vitamin C concentration during storage of orange juice.

[ 1 ]

Question 1

Question 1(a)

(a)

0.100 g of magnesium ribbon is added to 50.0 cm350.0 \mathrm{~cm}^{3} of 1.00moldm31.00 \mathrm{moldm}^{-3} sulfuric acid to produce hydrogen gas and magnesium sulfate.

Mg( s)+H2SO4(aq)H2( g)+MgSO4(aq)\mathrm{Mg}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{MgSO}_{4}(\mathrm{aq})
[ 2 ]

Question 1(a)(i)

(i)

The graph shows the volume of hydrogen produced against time under these experimental conditions.

Question image

Sketch two curves, labelled I and II, to show how the volume of hydrogen produced (under the same temperature and pressure) changes with time when:

I. using the same mass of magnesium powder instead of a piece of magnesium ribbon;
II. 0.100 g of magnesium ribbon is added to 50 cm350 \mathrm{~cm}^{3} of 0.500moldm30.500 \mathrm{moldm}^{-3} sulfuric acid.

[ 2 ]

Question 1

[Maximum number: 2]

This question is about a mug made of a lead alloy.

Question image

The rate of lead dissolving in common beverages with various pH values was analysed.

Lead dissolving in beverages at various times and temperatures

Lead dissolving in beverages at various times and temperatures

Question 1(a)

(a)

Identify the experiment with the highest rate of lead dissolving.

[ 1 ]

Question 1(b)

Question 1(b)(ii)

(b)
(i)

Examine, giving a reason, whether the rate of lead dissolving increases with acidity at 18C18^{\circ} \mathrm{C}.

[ 1 ]

Question 1

[Maximum number: 4]

Hydrogen peroxide decomposes to form water and oxygen.

2H2O2(aq)2H2O(l)+O2( g)2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})

The reaction is catalysed by solid manganese (IV) oxide, MnO2( s)\mathrm{MnO}_{2}(\mathrm{~s}).
A student carried out a series of experiments to determine how the rate of decomposition depends on the mass of catalyst. Each time a different mass of MnO2\mathrm{MnO}_{2} was added to 25.0 cm325.0 \mathrm{~cm}^{3} of hydrogen peroxide solution. The oxygen was collected in a graduated gas syringe and the volume recorded at regular intervals.

Figure 1

Figure 1

Question 1(b)

(a)

The student hypothesized, based on underlying theory, that doubling the mass of MnO2\mathrm{MnO}_{2} would double the rate of the catalysed reaction.

[ 2 ]

Question 1(b)(ii)

(i)

Explain how the student's hypothesis might be supported by collision theory.

[ 2 ]

Question 1(c)

(b)

The results from Figure 1 were processed to produce a graph showing how the initial rate varied with the mass of catalyst.

Figure 2

Figure 2

[ 2 ]

Question 1(c)(i)

(i)

Outline how the y-axis values on Figure 2 were obtained from the results in Figure 1.

[ 2 ]

Question 1

[Maximum number: 8]

This question is about the rate of reaction between bromine and methanoic acid.

Br2(aq)+HCOOH(aq)2Br(aq)+2H+(aq)+CO2( g)\mathrm{Br}_{2}(\mathrm{aq})+\mathrm{HCOOH}(\mathrm{aq}) \rightarrow 2 \mathrm{Br}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g})

Question 1(a)

(a)

State and explain how the rate of this reaction, measured in moldm3s1\mathbf{m o l} \mathbf{d m}^{-\mathbf{3}} \mathbf{s}^{-\mathbf{1}}, could be monitored experimentally.

[ 3 ]

Question 1(b)

(b)

The change in bromine concentration was monitored.

Question image
[ 5 ]

Question 1(b)(i)

(i)

Determine the instantaneous rate of reaction to two significant figures when [Br2]=0.0080 moldm3\left[\mathrm{Br}_{2}\right]=0.0080 \mathrm{~mol} \mathrm{dm}^{-3}.

[ 3 ]

Question 1(b)(ii)

(ii)

Outline why the graph has a negative non-linear slope.

Reason for negative slope:

Reason for non-linear slope:

[ 2 ]

Question 1

[Maximum number: 1]

In order to provide safe drinking water, a water supply is often treated with disinfectants, which aim to inactivate disease-causing bacteria in the water.

To compare the effectiveness of different disinfectants, a CT value is used as a measure of the dosage of disinfectant needed to achieve a certain level of inactivation of specific bacteria.

CT value (mgmindm3)=C(mgdm3)×T(min)\left(\mathrm{mg} \mathrm{min} \mathrm{dm}^{-3}\right)=\mathrm{C}\left(\mathrm{mg} \mathrm{dm}^{-3}\right) \times \mathrm{T}(\mathrm{min}) concentration contact time
of disinfectant with water

Question 1(b)

(a)

CT values are influenced by temperature and by pH . The table below shows the CT values for chlorine needed to achieve 99 % inactivation of a specific bacterium at stated values of pH and temperature.

Question image
[ 1 ]

Question 1(b)(i)

(i)

With reference to the temperature data in the table, suggest why it may be more difficult to treat water effectively with chlorine in cold climates.

[ 1 ]

Question 1

[Maximum number: 6]

Hydrogen peroxide, H2O2(aq)\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}), releases oxygen gas, O2( g)\mathrm{O}_{2}(\mathrm{~g}), as it decomposes according to the equation below.

2H2O2(aq)2H2O(l)+O2( g)2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})

50.0 cm350.0 \mathrm{~cm}^{3} of hydrogen peroxide solution was placed in a boiling tube, and a drop of liquid detergent was added to create a layer of bubbles on the top of the hydrogen peroxide solution as oxygen gas was released. The tube was placed in a water bath at 75C75^{\circ} \mathrm{C} and the height of the bubble layer was measured every thirty seconds. A graph was plotted of the height of the bubble layer against time.

Question image

Question 1(b)

(a)

Use the graph to calculate the rate of decomposition of hydrogen peroxide at 120 s .

[ 3 ]

Question 1(c)

(b)

The experiment was repeated using solid manganese(IV) oxide, MnO2( s)\mathrm{MnO}_{2}(\mathrm{~s}), as a catalyst.

[ 3 ]

Question 1(c)(i)

(i)

Draw a curve on the graph opposite to show how the height of the bubble layer changes with time when manganese(IV) oxide is present.

[ 1 ]

Question 1(c)(ii)

(ii)

Explain the effect of the catalyst on the rate of decomposition of hydrogen peroxide.

[ 2 ]
0 selected