Chemistry · General chemistry II · Worked example

Find the slow step on a two-step diagram

A reaction runs in two steps through an intermediate. Step 1 has a barrier of $50 kJ / mol$ and $Δ H_{1} = + 40 kJ / mol$ . Step 2 climbs $25 kJ / mol$ from the intermediate and has $Δ H_{2} = - 70 kJ / mol$ . Find the overall $Δ H$ , the height of each transition state above the reactants, the slow step, and the overall reverse barrier.

Δ H = Δ H_{1} + Δ H_{2}

Name the second step’s energy change

The first example had one hump. This path has two, with the intermediate in the valley between them. The first step’s energy change, $Δ H_{1} = + 40 kJ / mol$ , is the given, so the intermediate sits $40 kJ / mol$ above the reactants. The second step falls $70 kJ / mol$ from there.

Δ H_{2} = - 70 kJ/mol

The two steps $(kJ / mol)$
Step	Its own barrier	Its $Δ H$
1, reactants → intermediate	$50$	$+ 40$
2, intermediate → products	$25$	$- 70$

Add the steps

The overall reaction is the sum of the steps, so its $Δ H$ is the sum of theirs, the same rule as Hess’s law.

Δ H = Δ H_{1} + Δ H_{2}

Find the overall $Δ H$

$+ 40 + (- 70) = - 30 kJ / mol$ . The reaction is exothermic overall, although its first step is uphill.

Δ H = 40 kJ/mol + (- 70 kJ/mol) = - 30 kJ/mol

\begin{array}{c} Δ H = 40 + (- 70) kJ/mol \\ = - 30 kJ/mol \end{array}

Name the second barrier

Step 2’s own climb starts at the intermediate, not at the reactants.

E_{a, 2} = 25 kJ/mol

Write the height of the second top

Measured from the reactants, the second top is the intermediate’s level plus step 2’s climb.

E_{TS, 2} = Δ H_{1} + E_{a, 2}

Find it

$40 + 25 = 65 kJ / mol$ above the reactants. The first top is $50 kJ / mol$ above them, so the second top is the higher one, and step 2 is the slow step, though its own climb is only half of step 1’s.

E_{TS, 2} = 40 kJ/mol + 25 kJ/mol = 65 kJ/mol

\begin{array}{c} E_{TS, 2} = 40 + 25 kJ/mol \\ = 65 kJ/mol \end{array}

Write the reverse barrier

Run backward, the reaction climbs from the products to the highest top on the path.

E_{a, rev} = E_{TS, 2} - Δ H

Find it

$65 - (- 30) = 95 kJ / mol$ . The products sit $30 kJ / mol$ below the reactants and the highest top $65 kJ / mol$ above them.

E_{a, rev} = 65 kJ/mol - (- 30 kJ/mol) = 95 kJ/mol

\begin{array}{c} E_{a, rev} = 65 - (- 30) kJ/mol \\ = 95 kJ/mol \end{array}

Read the slow step and the intermediate

The intermediate is the valley at $+ 40 kJ / mol$ . When step 1 is fast and reversible, the climb the whole reaction must make is the highest top, $65 kJ / mol$ above the reactants, so step 2 sets the rate. This is the apparent activation energy an Arrhenius plot would give.

Result

$Δ H = - 30 kJ / mol$ . The tops are $50$ and $65 kJ / mol$ above the reactants, so step 2 is slow. The overall reverse barrier is $95 kJ / mol$ , and the intermediate sits at $+ 40 kJ / mol$ .

Your turn

1. One like the example

Step 1 has a barrier of $60 kJ / mol$ and $Δ H_{1} = + 15 kJ / mol$ . Step 2 climbs $35 kJ / mol$ from the intermediate and has $Δ H_{2} = - 45 kJ / mol$ . Find $Δ H$ , the height of the second top, and the slow step.

Show the answer and explanation

$Δ H = - 30 kJ / mol$ . The second top is $50 kJ / mol$ above the reactants, below the first at $60 kJ / mol$ , so step 1 is the slow step.

$Δ H = 15 + (- 45)$ $= - 30 kJ / mol$ . The intermediate sits at $+ 15$ , so the second top is $15 + 35 = 50 kJ / mol$ above the reactants. The first top, at $60 kJ / mol$ , is higher, so step 1 sets the rate.

\begin{array}{l} Δ H_{1} = 15 kJ/mol \\ Δ H_{2} = - 45 kJ/mol \\ Δ H = Δ H_{1} + Δ H_{2} \\ Δ H = 15 kJ/mol + (- 45 kJ/mol) = - 30 kJ/mol \\ E_{a, 2} = 35 kJ/mol \\ E_{TS, 2} = Δ H_{1} + E_{a, 2} \\ E_{TS, 2} = 15 kJ/mol + 35 kJ/mol = 50 kJ/mol \end{array}

\begin{array}{c} Δ H_{1} = 15 kJ/mol \\ Δ H_{2} = - 45 kJ/mol \\ Δ H = Δ H_{1} + Δ H_{2} \\ Δ H = 15 kJ/mol + (- 45 kJ/mol) \\ = - 30 kJ/mol \\ E_{a, 2} = 35 kJ/mol \\ E_{TS, 2} = Δ H_{1} + E_{a, 2} \\ E_{TS, 2} = 15 kJ/mol + 35 kJ/mol \\ = 50 kJ/mol \end{array}

2. A variation

An endothermic two-step reaction has $Δ H_{1} = + 30 kJ / mol$ and $Δ H_{2} = + 10 kJ / mol$ . Step 1’s barrier is $45 kJ / mol$ and step 2 climbs $50 kJ / mol$ from the intermediate. Find the second top, $Δ H$ , the slow step and the overall reverse barrier.

Show the answer and explanation

The second top is $80 kJ / mol$ above the reactants, so step 2 is slow. $Δ H = + 40 kJ / mol$ , and the reverse barrier is $40 kJ / mol$ .

Both steps are uphill this time. The second top is $30 + 50 = 80 kJ / mol$ , above the first at $45$ , so step 2 is slow. $Δ H = 30 + 10$ $= + 40 kJ / mol$ . Running backward, the reaction climbs from the products at $+ 40$ to the top at $80$ , a reverse barrier of $40 kJ / mol$ , smaller than the forward $80$ .

\begin{array}{l} Δ H_{1} = 30 kJ/mol \\ Δ H_{2} = 10 kJ/mol \\ E_{a, 2} = 50 kJ/mol \\ E_{TS, 2} = Δ H_{1} + E_{a, 2} \\ E_{TS, 2} = 30 kJ/mol + 50 kJ/mol = 80 kJ/mol \\ Δ H = Δ H_{1} + Δ H_{2} \\ Δ H = 30 kJ/mol + 10 kJ/mol = 40 kJ/mol \\ E_{a, rev} = E_{TS, 2} - Δ H \\ E_{a, rev} = 80 kJ/mol - 40 kJ/mol = 40 kJ/mol \end{array}

\begin{array}{c} Δ H_{1} = 30 kJ/mol \\ Δ H_{2} = 10 kJ/mol \\ E_{a, 2} = 50 kJ/mol \\ E_{TS, 2} = Δ H_{1} + E_{a, 2} \\ E_{TS, 2} = 30 kJ/mol + 50 kJ/mol \\ = 80 kJ/mol \\ Δ H = Δ H_{1} + Δ H_{2} \\ Δ H = 30 kJ/mol + 10 kJ/mol \\ = 40 kJ/mol \\ E_{a, rev} = E_{TS, 2} - Δ H \\ E_{a, rev} = 80 kJ/mol - 40 kJ/mol \\ = 40 kJ/mol \end{array}

3. Somewhere new

A catalyst lowers only the second top of the worked example by $20 kJ / mol$ . Which step is slow now, and what is the climb the whole reaction must make?

Show the answer and explanation

Step 1 becomes the slow step, and the reaction must climb $50 kJ / mol$ .

The second top falls from $65$ to $65 - 20 = 45 kJ / mol$ above the reactants. The first top is still at $50 kJ / mol$ , now the higher one, so step 1 sets the rate. The catalyst only helps until another step becomes the bottleneck. The overall climb fell from $65$ to $50 kJ / mol$ , not to $45$ .

\begin{array}{l} E_{TS, 2} = 65 kJ/mol \\ Δ E = 20 kJ/mol \\ E_{TS, 2, cat} = E_{TS, 2} - Δ E \\ E_{TS, 2, cat} = 65 kJ/mol - 20 kJ/mol = 45 kJ/mol \end{array}

\begin{array}{c} E_{TS, 2} = 65 kJ/mol \\ Δ E = 20 kJ/mol \\ E_{TS, 2, cat} = E_{TS, 2} - Δ E \\ E_{TS, 2, cat} = 65 kJ/mol - 20 kJ/mol \\ = 45 kJ/mol \end{array}

Keep exploring

Open the path in Energy diagrams & barriers and raise step 1’s barrier to $70 kJ / mol$ . Its top now stands at $70 kJ / mol$ , above the second, and step 1 becomes the rate-limiting step.

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After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 8 key terms Check ΔH and the barriers in a Chemistry box Draw the two-step path in Energy diagrams & barriers Open worked example on a board Chemistry formulas: kinetics

Your existing work stays on this device. Examples open as editable copies.

Name the second step’s energy change

Add the steps

Find the overall Δ⁢HΔH

Name the second barrier

Write the height of the second top

Find it

Write the reverse barrier

Find it

Read the slow step and the intermediate

Result

Your turn

1. One like the example

Δ⁢H=−30 kJ⁢/⁢molΔH=−30 kJ/mol. The second top is 50 kJ⁢/⁢mol50 kJ/mol above the reactants, below the first at 60 kJ⁢/⁢mol60 kJ/mol, so step 1 is the slow step.

2. A variation

The second top is 80 kJ⁢/⁢mol80 kJ/mol above the reactants, so step 2 is slow. Δ⁢H=+40 kJ⁢/⁢molΔH=+40 kJ/mol, and the reverse barrier is 40 kJ⁢/⁢mol40 kJ/mol.

3. Somewhere new

Step 1 becomes the slow step, and the reaction must climb 50 kJ⁢/⁢mol50 kJ/mol.

Keep exploring

Study it, then try it

Find the overall $Δ H$

$Δ H = - 30 kJ / mol$ . The second top is $50 kJ / mol$ above the reactants, below the first at $60 kJ / mol$ , so step 1 is the slow step.

The second top is $80 kJ / mol$ above the reactants, so step 2 is slow. $Δ H = + 40 kJ / mol$ , and the reverse barrier is $40 kJ / mol$ .

Step 1 becomes the slow step, and the reaction must climb $50 kJ / mol$ .