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
-
What is the minimum energy required by colliding particles for a successful chemical reaction?
- Ionisation energy
- Activation energy
- Enthalpy change
- Lattice energy
-
How does adding a catalyst increase the rate of a chemical reaction?
- Lowers enthalpy change
- Increases activation energy
- Increases enthalpy change
- Lowers activation energy
-
In the Arrhenius equation k = A e^(-Ea/RT), what does the symbol A represent?
- Rate constant
- Activation energy
- Pre-exponential factor
- Gas constant
-
A graph of ln k against 1/T is a straight line. What is its gradient?
- R/Ea
- Ea/R
- -A
- -Ea/R
-
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
-
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 increasing temperature increase the rate constant of a reaction?
- More successful collisions
- Lower activation energy
- Higher activation energy
- Fewer particle collisions
-
Two reactions have identical A values and temperatures. Which reaction has the larger rate constant?
- Higher enthalpy change
- Lower activation energy
- Lower temperature
- Higher activation energy
-
How does adding a catalyst affect a Maxwell-Boltzmann distribution curve?
- Flattens the peak
- Unchanged curve shape
- Shifts peak right
- Shifts peak left
-
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
-
In an Arrhenius plot of ln k against 1/T, how is activation energy calculated from the gradient?
- gradient / R
- -gradient / R
- -gradient × R
- gradient × R
-
Why does a graph of ln k against 1/T produce a negative gradient?
- k decreases with T
- Ea is negative
- Ea is positive
- 1/T increases with T
-
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 adding a catalyst not change the position of equilibrium?
- Increases product yield
- Speeds both directions
- Lowers reactant energy
- Only speeds forward
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