Rate Equation for HI Decomposition from k(T) Data: 23-Chem-A4 Chemical Reactor Engineering, December 2018, Q1

Question 1: Rate Equation for HI Decomposition from k(T) Data (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Rate-constant–temperature pairs for the gas-phase decomposition $2HI \rightarrow H_2 + I_2$. The units of $k$ (cm³·mol$^{-1}$·s$^{-1}$) are those of a second-order rate constant, so the elementary rate law is $-r_{HI} = k\,C_{HI}^{2}$. Temperatures are converted to kelvin ($T\,[\text{K}] = T\,[^\circ\text{C}] + 273.15$).

Find. The complete temperature-dependent rate expression — i.e. the Arrhenius pre-exponential $A$ and activation energy $E_a$ in $k = A\,e^{-E_a/RT}$.

Figure 1. Arrhenius plot. The five $(\,1/T,\ \ln k\,)$ points fall on a straight line ($R^2 = 0.999$); its slope is $-E_a/R$ and its intercept is $\ln A$.

Approach. Linearise the Arrhenius law as $\ln k = \ln A - (E_a/R)(1/T)$, fit a straight line to $\ln k$ versus $1/T$ by least squares, and read $E_a$ from the slope and $A$ from the intercept.

Tabulate the linearised variables. Convert each temperature to kelvin and form $1/T$ and $\ln k$. For example, at $283\,{}^\circ\text{C} = 556.15$ K, $\ln(9.42\times10^{-7}) = -13.88$; at $508\,{}^\circ\text{C} = 781.15$ K, $\ln(0.1059) = -2.245$. The five points span $1/T$ from $1.28\times10^{-3}$ to $1.80\times10^{-3}$ K$^{-1}$.
Least-squares fit. Regressing $\ln k$ on $1/T$ gives slope $=-2.241\times10^{4}$ K and intercept $=26.30$, with $R^2 = 0.9992$ — an excellent Arrhenius fit over the whole range.
Activation energy from the slope. With $R = 8.314\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$, $$E_a = -R\,(\text{slope}) = -(8.314)(-2.241\times10^{4}) = 1.863\times10^{5}\ \text{J/mol}.$$ $$\boxed{E_a = 186.3\ \text{kJ/mol}.}$$
Pre-exponential factor from the intercept. $A = e^{26.30} = 2.64\times10^{11}\ \text{cm}^3\,\text{mol}^{-1}\text{s}^{-1}$, carrying the same second-order units as the tabulated $k$.
Assemble the complete rate equation. Combining the Arrhenius constants with the second-order form gives $$-r_{HI} = \left(2.64\times10^{11}\ \tfrac{\text{cm}^3}{\text{mol}\cdot\text{s}}\right) \exp\!\left(\frac{-186{,}300}{RT}\right) C_{HI}^{2},$$ or equivalently, with $E_a/R = 2.241\times10^{4}$ K, $$\boxed{-r_{HI} = 2.64\times10^{11}\,e^{-22{,}410/T}\;C_{HI}^{2}\quad(T\text{ in K}).}$$

Quantity	Value
Activation energy $E_a$	186.3 kJ/mol
Pre-exponential $A$	$2.64\times10^{11}$ cm³·mol$^{-1}$s$^{-1}$
Correlation $R^2$	0.9992
Complete rate law	$-r_{HI} = 2.64\times10^{11}\,e^{-22410/T}\,C_{HI}^{2}$

Concept & Theory

Topic. Arrhenius temperature dependence and extraction of kinetic parameters from rate-constant data.

Key relations. Arrhenius law $k = A\,e^{-E_a/RT}$; linear form $\ln k = \ln A - (E_a/R)(1/T)$; the units of $k$ fix the reaction order (cm³·mol$^{-1}$s$^{-1}$ ⇒ second order).

Why this works. Taking logarithms turns the exponential into a straight line whose slope isolates $E_a$ independently of $A$. Because the collision/transition-state pre-factor varies only weakly with $T$ compared with the exponential, a two-parameter Arrhenius fit describes rate constants across hundreds of degrees with a single slope and intercept. The high $R^2$ confirms a single rate-controlling elementary step over 283–508 °C.

Common pitfall. Reporting $k$ without its units and thereby losing the order information; and using $T$ in °C instead of K in $1/T$ (which corrupts both slope and intercept).

Source. Levenspiel, Chemical Reaction Engineering, 3rd ed., Ch. 2 (kinetics of homogeneous reactions); Fogler, 5th ed., §3.3.

Question 1 of 5: Rate Equation for HI Decomposition from k(T) Data

Question 1: Rate Equation for HI Decomposition from k(T) Data (25 marks)