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A gas X breaks down by a first order reaction, so its rate constant has the unit s⁻¹…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A gas X breaks down by a first order reaction, so its rate constant has the unit $\text{s}^{-1}$ . Chemists use a plot like the one below to find the activation energy of a reaction without knowing anything about its mechanism. The rate constant $k$ is measured at several temperatures between $400\ \text{K}$ and $500\ \text{K}$ . The graph plots $\log_{10}(k / \text{s}^{-1})$ against $1000\ \text{K}/T$ , and the points fall on a straight line. Two points on the line are marked. At $1000\ \text{K}/T = 2.00$ (that is, $T = 500\ \text{K}$ ), $\log_{10}(k/\text{s}^{-1}) = -2.40$ . At $1000\ \text{K}/T = 2.50$ ( $T = 400\ \text{K}$ ), $\log_{10}(k/\text{s}^{-1}) = -5.90$ . Assume that the activation energy and the pre-exponential factor do not change with temperature. Take $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$ and $\ln 10 = 2.303$ . Notice that the vertical axis uses common logarithms, not natural ones. Give the activation energy in kJ per mole and the temperature in kelvin.

A graph of the common logarithm of the rate constant, with k in reciprocal seconds, against 1000 kelvin divided by the temperature. A straight line falls from left to right. Two points on it are marked: at 1000 K over T equal to 2.00 the logarithm is minus 2.40, and at 2.50 it is minus 5.90.

What is the activation energy of the reaction?
What is the pre-exponential factor $A$ ?
What is the rate constant at $450\ \text{K}$ ?
At what temperature is the half-life of X equal to $10.0\ \text{min}$ ?
By what factor does the rate constant grow when the temperature rises from $450\ \text{K}$ to $460\ \text{K}$ ?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Read the slope and the intercept of the line, and turn them into $E_a$ and $A$ . Then use the line for any other temperature.

(a) Slope. $\dfrac{-5.90 - (-2.40)}{2.50 - 2.00} = -7.00$ per unit of $1000\ \text{K}/T$ , so the slope against $1/T$ is $-7000\ \text{K}$ . Then

\displaystyle E_a = 2.303 \times 8.314 \times 7000\ \text{J mol}^{-1} = 134.0\ \text{kJ mol}^{-1}

(b) Intercept. $\log_{10} A = -2.40 + 7000 \times \dfrac{1}{500} = 11.60$ , so $A = 10^{11.60}\ \text{s}^{-1} = 3.98 \times 10^{11}\ \text{s}^{-1}$ .

(c) At $450\ \text{K}$ . $\log_{10} k = 11.60 - \dfrac{7000}{450} = 11.60 - 15.556 = -3.956$ , so $k = 1.11 \times 10^{-4}\ \text{s}^{-1}$ .

(d) A half-life of $10.0\ \text{min}$ . $k = \dfrac{0.693}{600\ \text{s}} = 1.155 \times 10^{-3}\ \text{s}^{-1}$ , and $\log_{10} k = -2.937$ . Then $\dfrac{1000\ \text{K}}{T} = \dfrac{11.60 - (-2.937)}{7.00} = 2.0768$ , so $T = 481.5\ \text{K}$ .

(e) From $450\ \text{K}$ to $460\ \text{K}$ . $\log_{10}\dfrac{k_2}{k_1} = 7000\left(\dfrac{1}{450} - \dfrac{1}{460}\right) = 0.3382$ , so the factor is $10^{0.3382} = 2.18$ .

Check. The same $E_a$ and $A$ , put into $k = Ae^{-E_a/RT}$ with natural logarithms, give back both marked points exactly.

Watch out. Reading the slope as if it were $-E_a/R$ (natural logarithms) gives only $58.2\ \text{kJ mol}^{-1}$ , smaller by the factor $2.303$ .

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