Chemistry · General chemistry II · Worked example

Find Ea from two rate constants

A reaction has $k = 2.5 \times 10^{- 3} s^{- 1}$ at $298 K$ and $k = 1.1 \times 10^{- 2} s^{- 1}$ at $318 K$ . Find $E_{a}$ , then predict $k$ at $338 K$ .

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

Name the second rate constant

The first example had four points and a fitted line. Here there are only two, so the two-point form replaces the fit. The first rate constant, $k_{1} = 2.5 \times 10^{- 3} s^{- 1}$ at $298 K$ , is the given. The second is $k_{2}$ at the higher temperature.

k_{2} = 1.1 \times 10^{- 2} s^{- 1}

Name the first temperature

$T_{1}$ goes with $k_{1}$ . Both temperatures are already in kelvin.

T_{1} = 298 K

Name the second temperature

$T_{2}$ goes with $k_{2}$ .

T_{2} = 318 K

Name $R$

$R$ in $J / (mol \cdot K)$ makes $E_{a}$ come out in $J / mol$ .

R = 8.314 J {mol}^{- 1} K^{- 1}

Write the two-point form

Writing $ln k = - E_{a} / R T + ln A$ at each temperature and subtracting cancels $ln A$ , which this data cannot give.

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

Solve for $E_{a}$ and put in the numbers

$k_{2} / k_{1} = 4.4$ , and $ln 4.4 = 1.482$ . The bracket is $1 / 298 - 1 / 318$ $= 2.111 \times 10^{- 4} K^{- 1}$ . $E_{a} = (8.314) (1.482)$ $\div (2.111 \times 10^{- 4})$ $= 5.84 \times 10^{4} J / mol$ .

E_{a} = \frac{(8.314) (1.482)}{2.111 \times 10^{- 4}} J/mol = 5.84 \times 10^{4} J/mol

\begin{array}{c} E_{a} = \frac{(8.314) (1.482)}{2.111 \times 10^{- 4}} J/mol \\ = 5.84 \times 10^{4} J/mol \end{array}

Convert to $kJ / mol$

Divide by $1000$ . $E_{a} = 58.4 kJ / mol$ , a typical barrier for a reaction that runs at a useful pace near room temperature.

E_{a} = 58.4 kJ/mol

Predict $k$ at $338 K$

Use the same equation with $T_{2} = 338 K$ and the $E_{a}$ just found. The exponent is $(58, 400 / 8.314) (1 / 298 - 1 / 338)$ $= 2.79$ , and $e^{2.79} = 16.3$ . Then $k = (2.5 \times 10^{- 3}) (16.3)$ $= 4.1 \times 10^{- 2} s^{- 1}$ . The two-point form gives no check that $E_{a}$ stays the same out to $338 K$ .

k_{338} = (2.5 \times 10^{- 3} s^{- 1}) e^{2.79} = 4.1 \times 10^{- 2} s^{- 1}

\begin{array}{c} k_{338} = (2.5 \times 10^{- 3} s^{- 1}) e^{2.79} \\ = 4.1 \times 10^{- 2} s^{- 1} \end{array}

Result

$E_{a} \approx 58.4 kJ / mol$ , and $k \approx 4.1 \times 10^{- 2} s^{- 1}$ at $338 K$ , an extrapolation with no check on curvature.

Your turn

1. One like the example

A reaction has $k = 1.00 \times 10^{- 3} s^{- 1}$ at $300. K$ and $k = 3.00 \times 10^{- 3} s^{- 1}$ at $320. K$ . What is $E_{a}$ ?

Show the answer and explanation

$43.8 kJ / mol$ .

$k$ tripled, so $ln (k_{2} / k_{1})$ $= ln 3.00$ $= 1.0986$ . The bracket is $1 / 300. - 1 / 320.$ $= 2.083 \times 10^{- 4} K^{- 1}$ . $E_{a} = 8.314 \times 1.0986$ $\div (2.083 \times 10^{- 4})$ $= 4.38 \times 10^{4} J / mol$ , or $43.8 kJ / mol$ .

\begin{array}{l} ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ k_{1} = 1.00 \times 10^{- 3} s^{- 1} \\ k_{2} = 3.00 \times 10^{- 3} s^{- 1} \\ T_{1} = 300. K \\ T_{2} = 320. K \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ E_{a} = \frac{(8.314 J {mol}^{- 1} K^{- 1}) ln (3.00)}{1 / (300. K) - 1 / (320. K)} = 4.38 \times 10^{4} J/mol \end{array}

\begin{array}{c} ln \frac{k_{2}}{k_{1}} \\ = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ k_{1} = 1.00 \times 10^{- 3} s^{- 1} \\ k_{2} = 3.00 \times 10^{- 3} s^{- 1} \\ T_{1} = 300. K \\ T_{2} = 320. K \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ E_{a} = \frac{8.314 ln (3.00)}{1 / 300. - 1 / 320.} J/mol \\ = 4.38 \times 10^{4} J/mol \end{array}

2. A variation

A reaction has $E_{a} = 50.0 kJ / mol$ . By what factor does $k$ rise when the temperature goes from $20.0 ° C$ to $30.0 ° C$ ?

Show the answer and explanation

By a factor of $1.97$ , almost exactly double.

This time $E_{a}$ is known and the ratio $k_{2} / k_{1}$ is the unknown. In kelvin the temperatures are $293.2 K$ and $303.2 K$ . The exponent is $(50, 000 \div 8.314) (1 / 293.2 - 1 / 303.2)$ $= 0.676$ , and $e^{0.676} = 1.97$ . This is where the rule of thumb that $10 K$ doubles a rate comes from.

\begin{array}{l} T_{1} = 20.0^{\circ} C + 273.15 = 293.2 K \\ T_{2} = 30.0^{\circ} C + 273.15 = 303.2 K \\ e^{\frac{5.00 \times 10^{4} J/mol}{8.314 J {mol}^{- 1} K^{- 1}} (\frac{1}{293.2 K} - \frac{1}{303.2 K})} = 1.97 \end{array}

\begin{array}{c} T_{1} = 20.0^{\circ} C + 273.15 = 293.2 K \\ T_{2} = 30.0^{\circ} C + 273.15 = 303.2 K \\ e^{\frac{5.00 \times 10^{4} J/mol}{8.314 J {mol}^{- 1} K^{- 1}} (\frac{1}{293.2 K} - \frac{1}{303.2 K})} \\ = 1.97 \end{array}

3. Somewhere new

A biochemist finds that an uncatalyzed reaction’s rate constant doubles, from $1.20 \times 10^{- 4} s^{- 1}$ to $2.40 \times 10^{- 4} s^{- 1}$ , between $25.0 ° C$ and $35.0 ° C$ . What $E_{a}$ does that imply?

Show the answer and explanation

About $53.0 kJ / mol$ .

Convert first, $298.2 K$ and $308.2 K$ . $ln (2.40 / 1.20)$ $= ln 2$ $= 0.693$ , and $1 / 298.2 - 1 / 308.2$ $= 1.088 \times 10^{- 4} K^{- 1}$ . $E_{a} = 8.314 \times 0.693$ $\div (1.088 \times 10^{- 4})$ $= 5.30 \times 10^{4} J / mol$ . A doubling over $10 K$ near room temperature always points to an $E_{a}$ near $53 kJ / mol$ .

\begin{array}{l} ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ k_{1} = 1.20 \times 10^{- 4} s^{- 1} \\ k_{2} = 2.40 \times 10^{- 4} s^{- 1} \\ T_{1} = 25.0^{\circ} C + 273.15 = 298.2 K \\ T_{2} = 35.0^{\circ} C + 273.15 = 308.2 K \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ E_{a} = \frac{(8.314 J {mol}^{- 1} K^{- 1}) ln (2.00)}{1 / (298.2 K) - 1 / (308.2 K)} = 5.30 \times 10^{4} J/mol \end{array}

\begin{array}{c} ln \frac{k_{2}}{k_{1}} \\ = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ k_{1} = 1.20 \times 10^{- 4} s^{- 1} \\ k_{2} = 2.40 \times 10^{- 4} s^{- 1} \\ T_{1} = 25.0^{\circ} C + 273.15 = 298.2 K \\ T_{2} = 35.0^{\circ} C + 273.15 = 308.2 K \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ E_{a} = \frac{8.314 ln (2.00)}{1 / 298.2 - 1 / 308.2} J/mol \\ = 5.30 \times 10^{4} J/mol \end{array}

Keep exploring

Enter both rate constants in Kinetics & Arrhenius and add $338 K$ as the prediction temperature. It predicts about $0.0406 s^{- 1}$ , the $4.1 \times 10^{- 2} s^{- 1}$ found here.

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Study the 6 key terms Check Ea in a Chemistry box Solve it in Kinetics & Arrhenius Open worked example on a board Chemistry formulas: experimental analysis

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Name the second rate constant

Name the first temperature

Name the second temperature

Name RR

Write the two-point form

Solve for EaEa​ and put in the numbers

Convert to kJ⁢/⁢molkJ/mol

Predict kk at 338 K338 K

Result

Your turn

1. One like the example

43.8 kJ⁢/⁢mol43.8 kJ/mol.

2. A variation

By a factor of 1.971.97, almost exactly double.

3. Somewhere new

About 53.0 kJ⁢/⁢mol53.0 kJ/mol.

Keep exploring

Study it, then try it

Name $R$

Solve for $E_{a}$ and put in the numbers

Convert to $kJ / mol$

Predict $k$ at $338 K$

$43.8 kJ / mol$ .

By a factor of $1.97$ , almost exactly double.

About $53.0 kJ / mol$ .