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Nitric oxide and chlorine form nitrosyl chloride: 2 NO(g) + Cl₂(g) → 2 NOCl(g)…

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

Nitric oxide and chlorine form nitrosyl chloride: $2\,\text{NO}(g) + \text{Cl}_2(g) \rightarrow 2\,\text{NOCl}(g)$ . Nitrosyl chloride is a yellow gas. The table gives initial rates of reaction measured at $300\ \text{K}$ , each found as $-\dfrac{d[\text{Cl}_2]}{dt}$ at the start of a run. The rate law has the form rate $= k[\text{NO}]^m[\text{Cl}_2]^n$ .

Run	$[\text{NO}]_0$ ( $\text{mol L}^{-1}$ )	$[\text{Cl}_2]_0$ ( $\text{mol L}^{-1}$ )	Initial rate ( $\text{mol L}^{-1}\ \text{s}^{-1}$ )
1	$0.010$	$0.020$	$2.40 \times 10^{-6}$
2	$0.030$	$0.010$	$1.08 \times 10^{-5}$
3	$0.020$	$0.040$	$1.92 \times 10^{-5}$

A proposed mechanism has two steps:

\displaystyle \text{NO} + \text{Cl}_2 \rightleftharpoons \text{NOCl}_2 \quad (\text{fast, equilibrium constant } K)

\displaystyle \text{NOCl}_2 + \text{NO} \rightarrow 2\,\text{NOCl} \quad (\text{slow, rate constant } k_2)

The figure shows its energy profile. The intermediate lies $8.0\ \text{kJ mol}^{-1}$ below the reactants, and the barrier of the slow step is $22.0\ \text{kJ mol}^{-1}$ above the intermediate. Treat this energy change of the fast step as its enthalpy change. The intermediate never builds up to a large concentration. Take $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$ .

An energy profile, not to scale, for a two-step mechanism. From the reactants, 2 NO plus Cl2, the energy rises over a small first peak and falls to a valley, the intermediate NOCl2 plus NO, marked 8.0 kJ per mole below the reactants. From the valley the energy rises over a higher second peak, marked 22.0 kJ per mole above the valley, and then falls far down to the products, 2 NOCl.

What is the order with respect to NO?
What is the order with respect to $\text{Cl}_2$ ?
What is the rate constant of the overall rate law at $300\ \text{K}$ ?
The mechanism gives rate $= k_2K[\text{NO}]^2[\text{Cl}_2]$ . What activation energy does an Arrhenius plot of the overall rate constant show?
By what factor does the overall rate constant grow from $300\ \text{K}$ to $320\ \text{K}$ ?

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. Find the orders from the table, then see how the mechanism ties the overall rate constant to the energy profile.

(a) and (b) Orders. Run 3 over run 1: $2^m \times 2^n = \dfrac{1.92 \times 10^{-5}}{2.40 \times 10^{-6}} = 8.0$ , so $m + n = 3$ . Run 2 over run 1: $3^m \times (0.5)^n = \dfrac{1.08 \times 10^{-5}}{2.40 \times 10^{-6}} = 4.5$ . With $n = 3 - m$ : $3^m \times 2^{m-3} = 4.5$ , so $6^m = 36$ and $m = 2$ , $n = 1$ . Check: $9 \times 0.5 = 4.5$ .

(c) Rate constant. From run 1, $k = \dfrac{2.40 \times 10^{-6}}{(0.010)^2 \times 0.020} = \dfrac{2.40 \times 10^{-6}}{2.0 \times 10^{-6}}\ \text{L}^2\ \text{mol}^{-2}\ \text{s}^{-1} = 1.20\ \text{L}^2\ \text{mol}^{-2}\ \text{s}^{-1}$ . Runs 2 and 3 give the same value.

The mechanism fits. rate $= k_2[\text{NOCl}_2][\text{NO}] = k_2K[\text{NO}]^2[\text{Cl}_2]$ : second order in NO and first order in $\text{Cl}_2$ , as measured. So $k = k_2K$ .

(d) Observed activation energy. $\ln k = \ln k_2 + \ln K$ . Against $1/T$ , $\ln k_2$ has slope $-E_2/R$ and $\ln K$ has slope $-\Delta H/R$ . So

\displaystyle E_{\text{obs}} = E_2 + \Delta H = 22.0 + (-8.0) = 14.0\ \text{kJ mol}^{-1}

This is the height of the top of the slow step above the reactants in the figure.

(e) From $300\ \text{K}$ to $320\ \text{K}$ . $\ln\dfrac{k_{320}}{k_{300}} = \dfrac{14000}{8.314}\left(\dfrac{1}{300} - \dfrac{1}{320}\right) = 0.3508$ , so the factor is $1.42$ .

Check. A least-squares fit of $\ln(\text{rate})$ against $\ln[\text{NO}]$ and $\ln[\text{Cl}_2]$ gives orders of exactly 2 and 1 and the same $k$ . Differentiating $\ln(k_2K)$ numerically gives the same $E_{\text{obs}}$ .

Watch out. The barrier of the slow step alone ( $22.0\ \text{kJ mol}^{-1}$ ) would give a factor of $1.74$ . Heating also shifts the fast equilibrium back towards NO and $\text{Cl}_2$ , which lowers $[\text{NOCl}_2]$ .

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