Unit 3 - Reaction rate

What is the average rate of the reaction over the first 5 minutes, in mol l^-1 min^-1?

• 3.6 × 10⁻⁴

• 8.4 × 10⁻³

• 8.4 × 10⁻⁴

• 1.2 × 10⁻³

Answer explanation

Average rate = change in quantity/ change in time

average rate = (0.0062-0.002)/5

average rate = 0.00084

In which of the following will both changes result in an increase in the rate of a chemical reaction?

• A decrease in activation energy and an increase in the frequency of collisions.

• An increase in activation energy and a decrease in particle size.

• An increase in temperature and an increase in the particle size.

• An increase in concentration and a decrease in the surface area of the reactant particles.

Graph X - 1g calcium carbonate powder with excess hydrochloric acid at 20°C

Which curve would best represent the reaction of a 0.5g lump of calcium carbonate with excess of the same dilute hydrochloric acid at at 20°C?

• A

• B

• C

• D

Answer explanation

Acid is in excess in both cases, so all calcium carbonate will be used up. 0.5g in A means less gas will be produced in comparison to X. Lump is used in A so reaction will be slower, as shown by less steep curve.

C and D produce same volume of gas as X so can rule out. B is faster than X so a lump was not used.

When a catalyst is used, the activation energy of the forward reaction is reduced to 35 kJ mol⁻¹. What is the activation energy of the catalysed reverse reaction?

• 30 kJ mol⁻¹

• 35 kJ mol⁻¹

• 65 kJ mol⁻¹

• 190 kJ mol⁻¹

Answer explanation

Reverse reaction is endothermic so answer is +.

Enthalpy change for backward reaction is 30, add this onto the activation energy of the reverse reaction = 65.

The graph shows how change in concentration of a reactant changes over time.

What is the average rate of this reaction, in mol l⁻¹ s⁻¹, between 10 and 20 s?

• 1.0 × 10⁻²

• 1.0 × 10⁻³

• 1.5 × 10⁻²

• 1.5 × 10⁻³

Answer explanation

average rate = change in quantity/ change in time

average rate = (0.035-0.025)/10

average rate = 0.001

In a reaction involving gases, an increase in temperature results in

• an increase in activation energy

• an increase in the enthalpy change

• a decrease in the activation energy

• more molecules per second forming an activated complex

Answer explanation

Increasing temperature gives reactant molecules more energy, so more of them will have an energy that is greater to or equal to the activation energy. This means they are more likely to collide successfully and form an activated complex.

In area X

• molecules always form an activated complex

• no molecules have the energy to form an activated complex

• collisions between molecules are always successful in forming products

• all molecules have the energy to form an activated complex

Answer explanation

The shaded area shows the number of molecules that have energy greater to or equal to activation energy.

This does not mean they will all form products. Remember, they also have to collide with the correct geometry.

When they collide successfully and form the activated complex, this unstable arrangement of atoms may break back down to reactants, so products are not always formed.

Which line in the table is correct for a reaction as the temperature decreases from T2 to T1?

• Activation energy remains the same, Number of successful collisions increases

• Activation energy decreases, Number of successful collisions decreases

• Activation energy decreases, Number of successful collisions increases

• Activation energy remains the same, Number of successful collisions decreases

Answer explanation

Activation energy only changes when a catalyst is added ( it decreases it). This is visually shown in the diagram as activation energy does not move.

When temperature decreases as in the question, less particles have energy equal to or greater than activation energy.

The following potential diagram is for a reaction carried out with and without a catalyst. The activation energy for the catalysed reaction is

• 30 kJ mol⁻¹

• 80 kJ mol⁻¹

• 100 kJ mol⁻¹

• 130 kJ mol⁻¹

Answer explanation

210-180 = 30. This gives distance between top of normal curve and dotted curve (catalysed).

130kjmol of energy is required for normal reaction. 130-30 = 100kjmol for catalysed reaction.

For any chemical, its temperature is a measure of

• the average kinetic energy of the particles that react

• the average kinetic energy of all the particles

• the activation energy

• the minimum kinetic energy required before reaction occurs