DP Chemistry: MC test: Activation energy

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet).

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

Which is the correct logarithmic form of the Arrhenius equation?

ln k = E_a/RT + ln A

ln k = − E_a/RT − ln A

ln k = − E_a/RT + ln A

ln k = E_a/RT − ln A

This is given in the IB data booklet but is useful to know for answering Paper 1 where the data booklet is not allowed.

For a first order reaction what are the units of A?

kJ mol⁻¹

s−1

It does not have units

dm³ mol⁻¹ s⁻¹

A has the same units as the rate constant.

What name is given to the constant A in the Arrhenius equation?

activation factor

pre-exponential factor

absolute temperature constant

activity constant

Although the correct name for A is the pre-exponential factor it is also known as the frequency constant. Perhaps not surprisingly it is also sometimes called the Arrhenius constant.

Which statements are correct about raising the temperature of a chemical reaction?

I. The natural logarithm of the rate constant becomes less negative.

II. A 10 K rise in temperature always doubles the rate of the reaction.

III. The average kinetic energy of the reactant particles increases.

I and III only

I and II only

II and III only

I, II and III

It is a rule of thumb that increasing the temperature by 10 K approximately doubles the rate but it is only true when the activation energy is about 50 kJ mol⁻¹ and the temperature rise is from about 298 K to 308 K. For example, if the activation energy is 100 kJ mol⁻¹ and the temperature is increased from 298 K to 308 K the rate nearly quadruples.

What does the constant A in the Arrhenius equation relate to?

the total number of collisions occurring

the number of collisions occurring which possess the activation energy to react

the total number of collisions with the average activation energy

the number of collisions occurring with the proper orientation to react

Many collisions do not result in a reaction. A is related to the orientation of the collisions as the reactant particles need to collide in a specific way as well as possess the necessary activation energy.

To find the activation energy for a reaction a graph of ln k against 1/T can be plotted to give a straight line where k is the rate constant and T is the absolute temperature.

Which will give the value for the activation energy directly?

taking the gradient

taking the intercept on the y axis

taking the gradient and multiplying by the gas constant

taking the gradient and dividing by the gas constant

The gradient = −E_a/R so to find E_a the gradient needs to be multiplied by R.

To find the pre-exponential factor, A in the Arrhenius equation a graph of ln k against 1/T can be plotted to give a straight line where k is the rate constant and T is the absolute temperature.

Which will give the value for the pre-exponential factor directly?

taking the antilogarithm of the value of the gradient

taking the antilogarithm of the value of the intercept on the y axis

the intercept on the y axis

the intercept on the x axis

The intercept on the y axis (ln k axis) is equal to ln A.

When the temperature of a reaction is increased from 500 K to 1000 K, what will be the value of ln (k₁/k₂) (where k₁ is the rate constant at 500 K and k₂ is the rate constant at 1000 K)?

E_a is the activation energy of the reaction and R is the gas constant.

E_a/0.001R

−E_a/0.003R

0.001 x E_a/R

−0.001 x E_a/R

ln k₁ = − E_a/RT₁ + ln A and ln k₂ = − E_a/RT₂+ ln A
so ln k₁ − ln k₂ = ln k₁/k₂ = − E_a/RT₁− (− E_a/RT₂) = E_a/R x (1/T₂− 1/T₁)
ln k₁/k₂ = E_a/R x (1/1000 − 1/500) = −0.001 x E_a/R

The activation energy of a reaction is equal to 60 kJ mol⁻¹. A catalyst provides an alternative pathway with an activation energy of 40 kJ mol⁻¹. What is the increase in the number of reaction particles with the necessary activation energy at 27 ^oC?

R is the gas constant.

e−^{(60000)/300R)} / e−^{(40000)/300R)}

e^{−(60000)/27R)} / e^{−(40000)/27R)}

e^{−(40000)/300R)} / e^{−(60000)/300R)}

e−^(60)/27R) / e^−(40)/27R)

At 27 ^oC without a catalyst the number of particles with an energy equal to or greater than E_a = e^{−(60 x 1000)/(R x 300)}
At 27 ^oC with a catalyst the number of particles with an energy equal to or greater than E_a = e^{−(40 x 1000)/(R x 300)}
The number has therefore increased by e^{−(40 x 1000)/(R x 300)} ÷ e^{−(60 x 1000)/(R x 300)}. E_a must be multiplied by 1000 as the units of R are J K⁻¹ mol⁻¹.

Which statements are correct concerning the Arrhenius equation?

I. A plot of ln k against 1/T will produce a straight line.

II. The activation energy, E_a is normally quoted in kJ mol⁻¹, whereas the gas constant, R is normally quoted in J K⁻¹ mol⁻¹.

III. The equation only applies to spontaneous reactions.

II and III only

I and III only

I and II only

I, II and III

The Arrhenius equation applies to all chemical reactions whether they are exothermic or endothermic. It will not work for some reactions above certain temperatures where one or more of the reactants decomposes before it can react, e.g. enzyme catalysed reactions.

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MC test: Activation energy

Multiple choice test on 16.2 Activation energy