**Using the Arrhenius Equation **

Rate constants for the gas-phase decomposition of hydrogen iodide, 2 HI(g) → H_2(g) + I_2(g), are listed in the following table:

(a) Find the activation energy (in kJ/mol) using all five data points.

(b) Calculate E_a from the rate constants at 283 °C and 508 °C.

(c) Given the rate constant at 283 °C and the value of E_a obtained in part (b), what is the rate constant at 293 °C?

**STRATEGY **

(a) The activation energy E_a can be determined from the slope of a linear plot of ln k versus 1/T.

(b) To calculate E_a from values of the rate constant at two temperatures, use the equation

\ln \left(\frac{k_2}{k_1}\right)=\left(\frac{-E_{\mathrm{a}}}{R}\right)\left(\frac{1}{T_2}-\frac{1}{T_1}\right)(c) Use the same equation and the known values of E_a and k_1 at T_1 to calculate k_2 at T_2.

Temperature (°C) | k (M^{-1} s^{-1} ) |

283 | 3.52 \times 10^{-7} |

356 | 3.02 \times 10^{-5} |

393 | 2.19 \times 10^{-4} |

427 | 1.16 \times 10^{-3} |

508 | 3.95 \times 10^{-2} |

Step-by-Step

Learn more on how do we answer questions.

(a) Because the temperature in the Arrhenius equation is expressed in kelvin, we must first convert the Celsius temperatures to absolute temperatures. Ten calculate values of 1/T and ln k, and plot ln k versus 1/T. The results are shown in the following table and graph:

The slope of the straight-line plot can be determined from the coordinates of any two widely separated points on the line:

\text { Slope }=\frac{\Delta y}{\Delta x}=\frac{(-14.0)-(-3.9)}{\left(0.00175 \mathrm{~K}^{-1}\right)-\left(0.00130 \mathrm{~K}^{-1}\right)}=\frac{-10.1}{0.00045 \mathrm{~K}^{-1}}=-2.24 \times 10^4 \mathrm{~K}Finally, calculate the activation energy from the slope:

\begin{aligned}E_{\mathrm{a}} & =-R(\text { Slope })=-\left(8.314 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\right)\left(-2.24 \times 10^4 \mathrm{~K}\right) \\& =1.9 \times 10^5 \mathrm{~J} / \mathrm{mol}=190 \mathrm{~kJ} / \mathrm{mol}\end{aligned}Note that the slope of the Arrhenius plot is negative and the activation energy is positive. The greater the activation energy for a particular reaction, the steeper the slope of the ln k versus 1/T plot and the greater the increase in the rate constant for a given increase in temperature.

(b) Substituting the values of k_1=3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1} at T_1=556 \mathrm{~K}\left(283^{\circ} \mathrm{C}\right) and k_2=3.95 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1} at T_2 = 781 K (508 °C) into the equation

\ln \left(\frac{k_2}{k_1}\right)=\left(\frac{-E_{\mathrm{a}}}{R}\right)\left(\frac{1}{T_2}-\frac{1}{T_1}\right)gives

\ln \left(\frac{3.95 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}}{3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1}}\right)=\left(\frac{-E_{\mathrm{a}}}{8.314 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}}\right)\left(\frac{1}{781 \mathrm{~K}}-\frac{1}{556 \mathrm{~K}}\right)Simplifying this equation gives

\begin{aligned}11.628 & =\left(\frac{-E_{\mathrm{a}}}{8.314 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}}\right)\left(\frac{-5.18 \times 10^{-4}}{\mathrm{~K}}\right) \\ E_{\mathrm{a}} & =1.87 \times 10^5 \mathrm{~J} / \mathrm{mol}=187 \ \mathrm{~kJ} / \mathrm{mol}\end{aligned}(c) Use the same equation as in part (b), but now the known values are

\begin{aligned}& k_1=3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1} \text { at } T_1=556 \mathrm{~K}\left(283^{\circ} \mathrm{C}\right) \\& E_{\mathrm{a}}=1.87 \times 10^5 \mathrm{~J} / \mathrm{mol}\end{aligned}and k_2 at T_2 = 566 K (293 °C) is the unknown.

\ln \left(\frac{k_2}{3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1}}\right)=\left(\frac{-1.87 \times 10^5 \frac{\mathrm{J}}{\mathrm{mol}}}{8.314 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}}\right)\left(\frac{1}{566 \mathrm{~K}}-\frac{1}{556 \mathrm{~K}}\right)=0.715Taking the antiln of both sides gives

\begin{aligned}& \frac{k_2}{3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1}}=e^{0.715}=2.04 \\& k_2=7.18 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1}\end{aligned}In this temperature range, a temperature increase of 10 K doubles the rate constant.

Temperature (°C) | T (K) | k (M^{-1} \ s^{-1} ) | 1/T (1/K) | ln k |

283 | 556 | 3.52 \times 10^{-7} | 0.001 80 | -14.86 |

356 | 629 | 3.02 \times 10^{-5} | 0.001 59 | -10.408 |

393 | 666 | 2.19 \times 10^{-4} | 0.001 50 | -8.426 |

427 | 700 | 1.16 \times 10^{-3} | 0.001 43 | -6.759 |

508 | 781 | 3.95 \times 10^{-2} | 0.001 28 | -3.231 |

Question: 12.11

Substituting the decay rates into the equation giv...

Question: 12.10

Substituting values for t and for t_{1/2}[/...

Question: 12.9

Substituting the values t_{1/2} = 3...

Question: 12.8

Substituting the values k = 4.63 × 10^{-2} ...

Question: 12.7

(a) Figure 12.1 shows that the concentration of [l...

Question: 12.6

Values of ln [ N_2O_5 ] are listed ...

Question: 12.5

(a) Because k has units of s^{-1} ,...

Question: 12.4

(a) Compare pairs of vessels in which the concentr...

Question: 12.3

(a) Comparing Experiments 1 and 2 shows that doubl...

Question: 12.2

The rate law for the second reaction in Table 12.2...