16.2 Activation Energy

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Question 1

Marks: 1

Which of the following statements about the constant A in the Arrhenius equation are correct?

It is a steric factor for the fraction of collisions where the particles have the correct mutual orientation
It takes into account the energy of the colliding particles
It takes into account the number of collisions in a chemical reaction

I and II only
I and III only
II and III only
I, II and III

Key Concepts

The Arrhenius Equation

Question 2

Marks: 1

The following information was obtained for the rate constant, k, for a reaction at 25 ^oC.

k	E_a	R
3.46 × 10^–8 s^–1	96.2 kJ mol^–1	8.31 J K^–1 mol^–1

Which expression correctly represents how to calculate the constant, A?

$A = \frac{(3.46 \times 10^{- 8})}{e^{(- 96.2 / 8.31 \times 25)}}$
$A = \frac{(3.46 \times 10^{- 8})}{e^{(- 96200 / 8.31 \times 298)}}$
$A = \frac{e^{(- 96200 / 8.31 \times 298)}}{(3.46 \times 10^{- 8})}$
$A = (3.46 \times 10^{- 8}) \times e^{(- 96200 / 8.31 \times 298)}$

Key Concepts

The Arrhenius Equation

Question 3

Marks: 1

ImoEHe96_example-arrhenius-plot

Which is the correct expression to calculate the activation energy? (R = 8.31J K^–1 mol^–1)

E_a = $\frac{1.90}{0.00020} \times (8.31 \times 10^{- 3})$
E_a = $\frac{- 1.90}{0.00020} \times 8.31$
E_a = $\frac{- 1.90}{0.00020} \times (8.31 \times 10^{- 3})$
E_a = $\frac{1.90}{0.00020} \times 8.31$

Key Concepts

Graphing the Arrhenius Equation

Question 4

Marks: 1

The rate constant data for a reaction at two different temperatures is shown.

Temperature / ^oC	Rate constant / mol^-1 dm³ s^-1
5	6.81 x 10^-6
35	6.11 x 10^-5

Using the following equation, which expression is the correct calculation for the activation energy of the reaction, in kJ mol^-1? (R = 8.31J K^–1 mol^–1)

$\ln \frac{k_{1}}{k_{2}} = \frac{E_{a}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$

$E_{a} = \frac{\ln \frac{6.11 \times 10^{- 5}}{6.81 \times 10^{- 6}} \times 8.31}{(\frac{1}{308} - \frac{1}{278})}$
$E_{a} = \frac{\ln \frac{6.81 \times 10^{- 6}}{6.11 \times 10^{- 5}} \times (8.31 \times 10^{- 3})}{(\frac{1}{278} - \frac{1}{308})}$
$E_{a} = \frac{\ln \frac{6.11 \times 10^{- 5}}{6.81 \times 10^{- 6}} \times 8.31}{(\frac{1}{278} - \frac{1}{308})}$
$E_{a} = \frac{\ln \frac{6.81 \times 10^{- 6}}{6.11 \times 10^{- 5}} \times (8.31 \times 10^{- 3})}{(\frac{1}{308} - \frac{1}{278})}$

Key Concepts

Activation Energy from Rate Constants at Different Temperatures

Question 5

Marks: 1

When the temperature increases from 25 ^oC to 55 ^oC, the rate constant for the reaction increases by a factor of 1.45.

$\ln \frac{k_{1}}{k_{2}} = \frac{E_{a}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$

Using the equation above, which expression is the correct calculation for the activation energy of the reaction? (R = 8.31J K^–1 mol^–1)

$E_{a} = \frac{(\ln 1.45) \times (8.31 \times 10^{- 3})}{(\frac{1}{55} - \frac{1}{25})}$
$E_{a} = \frac{(\ln 1.45) \times (8.31 \times 10^{- 3})}{(\frac{1}{328} - \frac{1}{298})}$
$E_{a} = \frac{(\ln \frac{1}{1.45}) \times (8.31 \times 10^{- 3})}{(\frac{1}{328} - \frac{1}{298})}$
$E_{a} = \frac{(\ln \frac{1.45}{1}) \times (8.31 \times 10^{- 3})}{(\frac{1}{55} - \frac{1}{25})}$

Key Concepts

Activation Energy from Rate Constants at Different Temperatures

DP IB Chemistry: HL

Topic Questions

16.2 Activation Energy

Question 1

Key Concepts

Question 2

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Question 3

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Question 4

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Question 5

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