Chemistry · General chemistry II · Worked example

Read a catalyzed potential energy diagram

A reaction has $Δ H = - 40 kJ / mol$ and $E_{a} = 50 kJ / mol$ . A catalyst lowers the forward barrier to $35 kJ / mol$ . Find the reverse barriers with and without the catalyst, say what happens to $Δ H$ and $K$ , and estimate how much faster the catalyzed reaction is at $298 K$ .

E_{a, rev} = E_{a} - Δ H

Place the levels

Put the reactants at $0$ . $Δ H$ is negative, so the products sit $40 kJ / mol$ below, and the reaction is exothermic. The uncatalyzed top is $50 kJ / mol$ above the reactants, and the catalyzed top is $35 kJ / mol$ above.

Levels relative to the reactants $(kJ / mol)$
Point	Uncatalyzed	Catalyzed
Reactants	$0$	$0$
Top of the hump	$+ 50$	$+ 35$
Products	$- 40$	$- 40$

Reverse barrier without the catalyst

The reverse reaction starts at the product level, $- 40 kJ / mol$ , and climbs to the same top at $+ 50$ . That climb is $50 - (- 40) = 90 kJ / mol$ .

E_{a, rev} = 50 kJ/mol - (- 40 kJ/mol) = 90 kJ/mol

\begin{array}{c} E_{a, rev} = 50 - (- 40) kJ/mol \\ = 90 kJ/mol \end{array}

Reverse barrier with the catalyst

The catalyzed top is at $+ 35$ , so the reverse climb is $35 - (- 40) = 75 kJ / mol$ . Both barriers fell by $15 kJ / mol$ .

E_{a, rev,cat} = 35 kJ/mol - (- 40 kJ/mol) = 75 kJ/mol

\begin{array}{c} E_{a, rev,cat} = 35 - (- 40) kJ/mol \\ = 75 kJ/mol \end{array}

What the catalyst leaves alone

The reactant and product levels did not move, so $Δ H$ is still $- 40 kJ / mol$ . The forward and reverse rate constants rise by the same factor, so their ratio, $K$ , is unchanged.

How much faster

With the same frequency factor $A$ , the rate constants differ by the Arrhenius factor for the $15 kJ / mol$ drop. $15, 000 \div (8.314 \times 298)$ $= 6.05$ , and $e^{6.05} = 426$ . Write the drop in $J / mol$ , because $R$ is in $J / (mol \cdot K)$ .

\frac{k_{cat}}{k} = e^{15000 / (8.314 \times 298)} \approx 4.3 \times 10^{2}

Result

The reverse barriers are $90 kJ / mol$ uncatalyzed and $75 kJ / mol$ catalyzed. $Δ H$ , $- 40 kJ / mol$ , and $K$ are unchanged. Both directions run about $430$ times faster at $298 K$ .

Your turn

1. One like the example

An endothermic reaction has $Δ H = + 30 kJ / mol$ and $E_{a} = 80 kJ / mol$ . What is the reverse activation energy, and could a catalyst lower the forward $E_{a}$ to $25 kJ / mol$ ?

Show the answer and explanation

The reverse $E_{a}$ is $50 kJ / mol$ . No catalyst could bring the forward $E_{a}$ down to $25 kJ / mol$ .

The products sit $30 kJ / mol$ above the reactants, and the top is $80 kJ / mol$ above them, so the reverse climb is $80 - 30 = 50 kJ / mol$ . The top of any path must lie above the products, which are already $30 kJ / mol$ up, so no forward barrier can be below $30 kJ / mol$ .

\begin{array}{l} Δ H = 30 kJ/mol \\ E_{a} = 80 kJ/mol \\ E_{a, rev} = E_{a} - Δ H \\ E_{a, rev} = 80 kJ/mol - 30 kJ/mol = 50 kJ/mol \end{array}

\begin{array}{c} Δ H = 30 kJ/mol \\ E_{a} = 80 kJ/mol \\ E_{a, rev} = E_{a} - Δ H \\ E_{a, rev} = 80 kJ/mol - 30 kJ/mol \\ = 50 kJ/mol \end{array}

2. A variation

On a diagram the reactants are at $60 kJ / mol$ , the top of the hump at $125 kJ / mol$ and the products at $- 20 kJ / mol$ . Find $E_{a}$ and $Δ H$ .

Show the answer and explanation

$E_{a} = 65 kJ / mol$ and $Δ H = - 80 kJ / mol$ , an exothermic reaction.

This time the levels are given, not the differences. $E_{a}$ is the climb from the reactant level, $125 - 60 = 65 kJ / mol$ , not the height of the top above the axis. $Δ H$ is the product level minus the reactant level, $- 20 - 60 = - 80 kJ / mol$ .

\begin{array}{l} H_{TS} = 125 kJ/mol \\ H_{R} = 60 kJ/mol \\ H_{P} = - 20 kJ/mol \\ E_{a} = H_{TS} - H_{R} \\ E_{a} = 125 kJ/mol - 60 kJ/mol = 65 kJ/mol \\ Δ H = H_{P} - H_{R} \\ Δ H = - 20 kJ/mol - 60 kJ/mol = - 80 kJ/mol \end{array}

\begin{array}{c} H_{TS} = 125 kJ/mol \\ H_{R} = 60 kJ/mol \\ H_{P} = - 20 kJ/mol \\ E_{a} = H_{TS} - H_{R} \\ E_{a} = 125 kJ/mol - 60 kJ/mol \\ = 65 kJ/mol \\ Δ H = H_{P} - H_{R} \\ Δ H = - 20 kJ/mol - 60 kJ/mol \\ = - 80 kJ/mol \end{array}

3. Somewhere new

An enzyme lowers a reaction’s barrier by $10.0 kJ / mol$ at body temperature, $310 K$ . With the same frequency factor, by what factor does it speed the reaction up?

Show the answer and explanation

About $48$ times faster.

The ratio of rate constants is $e^{Δ E_{a} / R T}$ . $10, 000 J / mol$ $\div (8.314 J / (mol \cdot K) \times 310 K)$ $= 3.88$ , and $e^{3.88} = 48.4$ . Real enzymes lower barriers by far more than $10 kJ / mol$ , which is how they speed reactions up by factors of millions.

\begin{array}{l} T = 310 K \\ e^{\frac{10000 J/mol}{(8.314 J {mol}^{- 1} K^{- 1}) (310 K)}} = 48.4 \end{array}

Keep exploring

Open the diagram in Energy diagrams & barriers and set the catalyzed barrier to $25 kJ / mol$ . The reverse barrier with the catalyst falls to $65 kJ / mol$ , while $Δ H$ stays at $- 40 kJ / mol$ .

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Place the levels

Reverse barrier without the catalyst

Reverse barrier with the catalyst

What the catalyst leaves alone

How much faster

Result

Your turn

1. One like the example

The reverse EaEa​ is 50 kJ⁢/⁢mol50 kJ/mol. No catalyst could bring the forward EaEa​ down to 25 kJ⁢/⁢mol25 kJ/mol.

2. A variation

Ea=65 kJ⁢/⁢molEa​=65 kJ/mol and Δ⁢H=−80 kJ⁢/⁢molΔH=−80 kJ/mol, an exothermic reaction.

3. Somewhere new

About 4848 times faster.

Keep exploring

Study it, then try it

The reverse $E_{a}$ is $50 kJ / mol$ . No catalyst could bring the forward $E_{a}$ down to $25 kJ / mol$ .

$E_{a} = 65 kJ / mol$ and $Δ H = - 80 kJ / mol$ , an exothermic reaction.

About $48$ times faster.