Lesson 16.1.5
16.1.5 Activation energy from rate data Quiz: Pearson Edexcel Chemistry, Unit 16
20 questions
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Lesson 16.1.5, Activation energy from rate data: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 16: Kinetics II, written with Revision Ninja.
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The 20 questions
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What does the activation energy represent?
- The minimum energy that colliding particles must have for a collision to lead to reaction
- The total energy released when the products are formed from the reactants
- The average kinetic energy of all the particles in the reaction mixture
- The energy required to break all bonds in one mole of the reactant
-
What effect does a catalyst have on activation energy?
- It increases the enthalpy change of the reaction, so the products are more stable than before the catalyst was added
- It lowers the activation energy by providing an alternative pathway, without changing the enthalpy change of the reaction
- It has no effect on the activation energy, but increases the number of collisions per second in the mixture
- It raises the activation energy for the forward reaction only, so fewer particles can react at the same temperature
-
In the Arrhenius equation k = A e^(-Ea/RT), what does A represent?
- The temperature in kelvin at which the reaction starts, which is set by the external heat source in the lab
- The activation energy in joules per mole, which is the energy barrier that colliding particles must overcome
- The gas constant, which has a value of 8.31 J K-1 mol-1 and appears in the exponential term of the equation
- The pre-exponential factor, which relates to the frequency of collisions with the correct orientation
-
A graph of ln k against 1/T is a straight line. What is its gradient?
- R/Ea
- Ea/R
- -A
- -Ea/R
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The Arrhenius equation is used with R in J K-1 mol-1 and Ea in J mol-1. What is the unit of Ea/R?
- kJ mol-1
- K
- J mol-1 K
- mol K J-1
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Which rearrangement of the Arrhenius equation gives y = mx + c with y = ln k and x = 1/T?
- ln k = ln A - (Ea/R)(1/T)
- k = ln A - Ea/RT
- ln k = Ea/RT - ln A
- ln k = ln A + (Ea/R)(1/T)
-
Why does raising the temperature increase the rate of a reaction?
- The total number of particles in the vessel increases as the temperature rises, so there are more collisions
- The activation energy falls as the temperature rises, so the barrier is lower and every collision succeeds
- More particles have energy at or above the activation energy, so a greater fraction of collisions are successful
- The enthalpy change becomes more negative as the temperature rises, so the reaction releases energy faster
-
Two reactions have the same pre-exponential factor. At the same temperature, which has the larger rate constant?
- The reaction with the lower activation energy, since more molecules exceed the lower energy barrier
- The reaction with the higher enthalpy change, since it releases more energy
- Both have the same rate constant, since A is the same
- The reaction with the higher activation energy, since more collisions have the energy needed
-
On a Maxwell-Boltzmann distribution, what does adding a catalyst do?
- It increases the number of particles with energy greater than the activation energy, without changing the shape of the curve
- It moves the peak of the curve to a higher energy and increases the total area under the whole distribution curve
- It lowers the peak energy of the curve and reduces the total number of particles present in the reaction mixture
- It makes the curve steeper at the peak while keeping the activation energy line in the same position on the graph
-
A plot of ln k against 1/T has a gradient of -4.8 x 10^3 K. What is the activation energy?
- 580 kJ mol-1
- 40 kJ mol-1
- 4.8 kJ mol-1
- -40 kJ mol-1
-
The rate constant doubles when temperature rises from 298 K to 308 K, approximately. Which activation energy best fits?
- About 53 kJ mol-1
- About 530 kJ mol-1
- About 5.3 kJ mol-1
- About 106 kJ mol-1
-
A reaction has Ea = 75 kJ mol-1. By what factor does k increase when temperature rises from 298 K to 308 K?
- About 10
- About 1.5
- About 2.7
- About 5.4
-
Ea = 50 kJ mol-1. By what factor does k increase when temperature rises from 300 K to 320 K?
- About 3.5
- About 10
- About 7.0
- About 1.3
-
A catalyst lowers Ea from 80 kJ mol-1 to 50 kJ mol-1. By what factor does the rate constant increase at 300 K, assuming A is unchanged?
- About 1.6
- About 1.7 x 10^3
- About 30
- About 1.7 x 10^5
-
Points on a ln k against 1/T plot are (3.00 x 10^-3 K-1, -10.0) and (2.80 x 10^-3 K-1, -6.0). What is Ea?
- About 166 kJ mol-1
- About 20 kJ mol-1
- About 83 kJ mol-1
- About 2.0 kJ mol-1
-
A student reports Ea = -45 kJ mol-1 from a plot. Which evaluation is best?
- The value cannot be right, since Ea is always positive, so the sign of the gradient has been mishandled
- The value is right, since the plot had a positive gradient for an endothermic reaction
- The value is right, since catalysts make the activation energy negative
- The value is right, since a negative activation energy shows the reaction is exothermic
-
Why does a plot of ln k against 1/T have a negative gradient?
- The pre-exponential factor A is negative for all exothermic reactions
- The gas constant R is negative, which makes ln k fall as temperature rises
- The Arrhenius equation includes a negative sign that always reverses the gradient
- k rises with T, so ln k rises as 1/T falls, which gives a negative gradient
-
What is the value of the gas constant R used in the Arrhenius equation?
- 9.81 m s-2
- 8.31 J K-1 mol-1
- 1.38 x 10^-23 J K-1
- 6.02 x 10^23 mol-1
-
Ea = 60 kJ mol-1. By what factor does k increase when temperature rises from 300 K to 310 K?
- About 1.3
- About 10
- About 4.5
- About 2.2
-
Why does a catalyst not change the position of an equilibrium?
- It increases the enthalpy change of the reaction, which favours the products and so shifts the equilibrium to the right
- It raises the temperature of the mixture locally, which shifts the equilibrium towards the endothermic side of the reaction
- It removes products as they form, which pulls the equilibrium forward and so changes the final position of the mixture
- It lowers the activation energy of the forward and reverse reactions equally, so equilibrium is reached faster at the same position
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