Rate laws and order of reaction
In the lab you add the same extra amount of reactant to two reactions. In one the rate doubles, in another it quadruples, and in a third it doesn't change at all. For $2\,\mathrm{NO} + \mathrm{O_2} \rightarrow 2\,\mathrm{NO_2}$ it is tempting to write $v = k[\mathrm{NO}]^2[\mathrm{O_2}]$ just by copying the coefficients. Here that happens to be right, but only because experiments say so. How do we find out the true dependence, and what does it tell us about how the reaction happens?
The rate law and partial orders
In the previous section you saw that concentration affects the rate of reaction. Here we put numbers on that effect. Different reactions respond very differently to the same change in concentration, so we need a way of writing that sensitivity mathematically.
For the gas-phase reaction $2\,\mathrm{NO}\,(g) + \mathrm{O_2}\,(g) \rightarrow 2\,\mathrm{NO_2}\,(g)$, it has been determined experimentally that its rate equation at…
For the gas-phase reaction $2\,\mathrm{NO}\,(g) + \mathrm{O_2}\,(g) \rightarrow 2\,\mathrm{NO_2}\,(g)$, it has been determined experimentally that its rate equation at $300\,\mathrm{K}$ is $v = k\,[\mathrm{NO}]^2\,[\mathrm{O_2}]$. Justify, with reasoning: a) Why the partial orders with respect to each reactant cannot be deduced directly from the stoichiometric coefficients of the overall chemical equation. b) What the overall order of the reaction is and what the units of the rate constant $k$ are, expressing time in seconds and concentrations in $\mathrm{mol\times L^{-1}}$.
The kinetics of the reaction $\mathrm{CH_3Br}\,(aq) + \mathrm{OH^-}\,(aq) \rightarrow \mathrm{CH_3OH}\,(aq) + \mathrm{Br^-}\,(aq)$ at $298\,\mathrm{K}$ is studied. The…
The kinetics of the reaction $\mathrm{CH_3Br}\,(aq) + \mathrm{OH^-}\,(aq) \rightarrow \mathrm{CH_3OH}\,(aq) + \mathrm{Br^-}\,(aq)$ at $298\,\mathrm{K}$ is studied. The process is known to be first order with respect to bromomethane and first order with respect to the hydroxide ion, with an initial rate of $1.2\times 10^{-5}\,\mathrm{mol\times L^{-1}\times s^{-1}}$ when $[\mathrm{CH_3Br}]_0 = 0.020\,\mathrm{mol\times L^{-1}}$ and $[\mathrm{OH^-}]_0 = 0.030\,\mathrm{mol\times L^{-1}}$. a) Write the rate law and calculate the numerical value of the rate constant $k$, stating its units. b) Calculate the rate if $[\mathrm{CH_3Br}]$ is doubled and $[\mathrm{OH^-}]$ is halved.
Zero order, first order and pseudo-order
Here $x$ is the concentration and the vertical axis is the rate. Compare the three curves. Zero order is a horizontal line, so $v$ does not change with concentration. First order is a straight line through the origin. Second order is a curve that rises faster and faster.
The decomposition of the gaseous reactant $\mathrm{A}$ follows zero-order kinetics with respect to its concentration at a constant temperature of $500\,\mathrm{K}$, with…
The decomposition of the gaseous reactant $\mathrm{A}$ follows zero-order kinetics with respect to its concentration at a constant temperature of $500\,\mathrm{K}$, with a rate constant $k = 4.5\times 10^{-4}\,\mathrm{mol\times L^{-1}\times s^{-1}}$. a) Write the differential rate equation for this process and deduce the units of its rate constant. b) If the initial concentration of $\mathrm{A}$ is tripled, state, with reasoning, how the numerical value of the instantaneous reaction rate changes.
The kinetics of the reaction $A + B \rightarrow Productos$ at $300\,\mathrm{K}$ is studied under reactant-isolation (pseudo-order) conditions. In a first series with a…
The kinetics of the reaction $A + B \rightarrow Productos$ at $300\,\mathrm{K}$ is studied under reactant-isolation (pseudo-order) conditions. In a first series with a large excess of $B$ ($[B]_0 = 2.0\,\mathrm{mol\times L^{-1}}$ constant), the initial rate is found to be proportional to $[A]_0^2$. In a second series with a large excess of $A$ ($[A]_0 = 1.5\,\mathrm{mol\times L^{-1}}$ constant), the initial rate does not change when $[B]_0$ is doubled or tripled. a) Formulate the complete differential rate law, justifying the partial orders and the overall order, and state the units of the rate constant. b) If in an experiment with $[A]_0 = 0.10\,\mathrm{mol\times L^{-1}}$ and $[B]_0 = 0.20\,\mathrm{mol\times L^{-1}}$ the initial rate is $v_0 = 3.5\times 10^{-3}\,\mathrm{mol\times L^{-1}\times s^{-1}}$, calculate the rate that would be recorded if the solution is diluted to twice its volume by adding pure solvent at the same temperature.
The method of initial rates
For the reaction $2\,\mathrm{A}\,(g) + \mathrm{B}\,(g) \rightarrow \mathrm{C}\,(g)$ at $350\,\mathrm{K}$, the initial rate is measured while keeping the concentration…
For the reaction $2\,\mathrm{A}\,(g) + \mathrm{B}\,(g) \rightarrow \mathrm{C}\,(g)$ at $350\,\mathrm{K}$, the initial rate is measured while keeping the concentration $[\mathrm{B}] = 0.40\,\mathrm{mol\times L^{-1}}$ constant and recording the response as $[\mathrm{A}]$ is varied. The graph of the experimental data obtained is shown below:
In the experiment, the pairs of values $([\mathrm{A}] = 0.10\,\mathrm{mol\times L^{-1}};\, v_0 = 0.0020\,\mathrm{mol\times L^{-1}\times s^{-1}})$, $([\mathrm{A}] = 0.20\,\mathrm{mol\times L^{-1}};\, v_0 = 0.0080\,\mathrm{mol\times L^{-1}\times s^{-1}})$ and $([\mathrm{A}] = 0.30\,\mathrm{mol\times L^{-1}};\, v_0 = 0.0180\,\mathrm{mol\times L^{-1}\times s^{-1}})$ were obtained. a) From the graph or the data provided, deduce the partial order with respect to $\mathrm{A}$. b) If an additional experiment at $[\mathrm{A}] = 0.10\,\mathrm{mol\times L^{-1}}$ and $[\mathrm{B}] = 0.80\,\mathrm{mol\times L^{-1}}$ gives $v_0 = 0.0040\,\mathrm{mol\times L^{-1}\times s^{-1}}$, obtain the complete rate equation and calculate the constant $k$.
For the reaction in aqueous solution $\mathrm{S_2O_8^{2-}} + 2\,\mathrm{I^-} \rightarrow 2\,\mathrm{SO_4^{2-}} + \mathrm{I_2}$, the following initial rate data were…
For the reaction in aqueous solution $\mathrm{S_2O_8^{2-}} + 2\,\mathrm{I^-} \rightarrow 2\,\mathrm{SO_4^{2-}} + \mathrm{I_2}$, the following initial rate data were obtained at $298\,\mathrm{K}$:
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Experiment 1: $[\mathrm{S_2O_8^{2-}}]_0 = 0.080\,\mathrm{mol\times L^{-1}}$, $[\mathrm{I^-}]_0 = 0.034\,\mathrm{mol\times L^{-1}}$, $v_0 = 2.2\times 10^{-4}\,\mathrm{mol\times L^{-1}\times s^{-1}}$.
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Experiment 2: $[\mathrm{S_2O_8^{2-}}]_0 = 0.080\,\mathrm{mol\times L^{-1}}$, $[\mathrm{I^-}]_0 = 0.017\,\mathrm{mol\times L^{-1}}$, $v_0 = 1.1\times 10^{-4}\,\mathrm{mol\times L^{-1}\times s^{-1}}$.
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Experiment 3: $[\mathrm{S_2O_8^{2-}}]_0 = 0.160\,\mathrm{mol\times L^{-1}}$, $[\mathrm{I^-}]_0 = 0.017\,\mathrm{mol\times L^{-1}}$, $v_0 = 2.2\times 10^{-4}\,\mathrm{mol\times L^{-1}\times s^{-1}}$.
- a) Determine the partial orders with respect to each reactant and the overall order. b) Calculate the value of the rate constant with its corresponding units and predict the initial rate if $[\mathrm{S_2O_8^{2-}}]_0 = 0.040\,\mathrm{mol\times L^{-1}}$ and $[\mathrm{I^-}]_0 = 0.050\,\mathrm{mol\times L^{-1}}$.
Several reactants changed at once
Squeeze a gas mixture into half the volume and the concentration of every gas doubles at the same time. How do simultaneous changes in several reactants combine to affect the rate?
Consider the gas-phase synthesis reaction $\mathrm{X}\,(g) + 2\,\mathrm{Y}\,(g) \rightarrow \mathrm{Z}\,(g)$, whose empirical rate equation is…
Consider the gas-phase synthesis reaction $\mathrm{X}\,(g) + 2\,\mathrm{Y}\,(g) \rightarrow \mathrm{Z}\,(g)$, whose empirical rate equation is $v = k\,[\mathrm{X}]\,[\mathrm{Y}]^2$. In a fixed-volume reactor $V_1$ at temperature $T$, the reaction proceeds with an initial rate $v_1$. a) Explain quantitatively how the initial rate changes if, keeping the temperature and volume constant, the amount of $\mathrm{X}$ is tripled and the amount of $\mathrm{Y}$ is halved. b) If the initial mixture is transferred to a vessel of volume $V_2 = V_1 / 3$ at the same temperature, calculate the factor by which the reaction rate is multiplied relative to the original rate in $V_1$.
The reaction $2\,\mathrm{NO}\,(g) + 2\,\mathrm{H_2}\,(g) \rightarrow \mathrm{N_2}\,(g) + 2\,\mathrm{H_2O}\,(g)$ at $1090\,\mathrm{K}$ is studied kinetically using the…
The reaction $2\,\mathrm{NO}\,(g) + 2\,\mathrm{H_2}\,(g) \rightarrow \mathrm{N_2}\,(g) + 2\,\mathrm{H_2O}\,(g)$ at $1090\,\mathrm{K}$ is studied kinetically using the method of initial rates. The laboratory data, with the experimental measurement uncertainty, are given below:
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Experiment 1: $[\mathrm{NO}]_0 = (4.0 \pm 0.1)\times 10^{-3}\,\mathrm{mol\times L^{-1}}$, $[\mathrm{H_2}]_0 = (1.5 \pm 0.1)\times 10^{-3}\,\mathrm{mol\times L^{-1}}$, $v_0 = (3.0 \pm 0.1)\times 10^{-5}\,\mathrm{mol\times L^{-1}\times s^{-1}}$.
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Experiment 2: $[\mathrm{NO}]_0 = (4.0 \pm 0.1)\times 10^{-3}\,\mathrm{mol\times L^{-1}}$, $[\mathrm{H_2}]_0 = (3.0 \pm 0.1)\times 10^{-3}\,\mathrm{mol\times L^{-1}}$, $v_0 = (6.0 \pm 0.2)\times 10^{-5}\,\mathrm{mol\times L^{-1}\times s^{-1}}$.
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Experiment 3: $[\mathrm{NO}]_0 = (8.0 \pm 0.1)\times 10^{-3}\,\mathrm{mol\times L^{-1}}$, $[\mathrm{H_2}]_0 = (1.5 \pm 0.1)\times 10^{-3}\,\mathrm{mol\times L^{-1}}$, $v_0 = (1.2 \pm 0.1)\times 10^{-4}\,\mathrm{mol\times L^{-1}\times s^{-1}}$.
- a) Deduce the rate law, the overall order and the mean value of $k$ with its units. b) The following mechanism is proposed:
Step 1 (fast, equilibrium): $2\,\mathrm{NO} \rightleftharpoons \mathrm{N_2O_2}$
Step 2 (slow): $\mathrm{N_2O_2} + \mathrm{H_2} \rightarrow \mathrm{N_2O} + \mathrm{H_2O}$
Step 3 (fast): $\mathrm{N_2O} + \mathrm{H_2} \rightarrow \mathrm{N_2} + \mathrm{H_2O}$
Justify whether this mechanism is kinetically consistent with the rate law deduced.
Log graphs and reaction mechanisms
Finding the order from a graph
Taking natural logs of $v_0 = k[\mathrm{A}]^n$ (other concentrations constant) gives a straight line: $$\ln v_0 = \ln k + n\ln[\mathrm{A}]_0$$ If you plot $\ln v_0$ on the y-axis against $\ln[\mathrm{A}]_0$ on the x-axis, the gradient is the order $n$ and the y-intercept is $\ln k$. Then $k = e^{\text{intercept}}$.
When data have measurement uncertainty, the gradient comes out as something like $1.98 \pm 0.05$. Give the nearest whole (or half) number that fits within the uncertainty and say why.
What the rate law tells you about the mechanism
Most reactions happen in several elementary steps, called the mechanism. The molecularity of an elementary step is the number of particles colliding in it, and for an elementary step the rate law does follow the coefficients. In a mechanism, the slowest step is the rate-determining step and sets the rate law. The fast steps after it do not appear.
The gas-phase thermal decomposition reaction $2\,\mathrm{N_2O_5}\,(g) \rightarrow 4\,\mathrm{NO_2}\,(g) + \mathrm{O_2}\,(g)$ at $318\,\mathrm{K}$ is analysed. The plot…
The gas-phase thermal decomposition reaction $2\,\mathrm{N_2O_5}\,(g) \rightarrow 4\,\mathrm{NO_2}\,(g) + \mathrm{O_2}\,(g)$ at $318\,\mathrm{K}$ is analysed. The plot of $\ln v_0$ against $\ln [\mathrm{N_2O_5}]_0$ for four independent experiments is observed to give a straight line of gradient $1.00 \pm 0.02$ with a y-intercept equal to $-6,215$. a) Determine, with justification, the order of reaction and the value of the rate constant $k$ with its corresponding units. b) Consider a proposed mechanism consisting of three elementary steps:
1) $\mathrm{N_2O_5} \rightleftharpoons \mathrm{NO_2} + \mathrm{NO_3}$ (fast equilibrium, constants $k_1$ and $k_{-1}$)
2) $\mathrm{NO_2} + \mathrm{NO_3} \rightarrow \mathrm{NO_2} + \mathrm{O_2} + \mathrm{NO}$ (slow step, constant $k_2$)
3) $\mathrm{NO} + \mathrm{NO_3} \rightarrow 2\,\mathrm{NO_2}$ (fast step, constant $k_3$)
Deduce analytically the expression for the rate law predicted by this mechanism and check whether it reproduces the experimental order. The logarithms are applied to the numerical values of the rates in $\mathrm{mol}\times\mathrm{L}^{-1}\times\mathrm{s}^{-1}$ and of the concentrations in $\mathrm{mol}\times\mathrm{L}^{-1}$, taking those units as references.
Back to the start: the coefficients in $2\,\mathrm{NO} + \mathrm{O_2} \rightarrow 2\,\mathrm{NO_2}$ give no guarantee of the orders. Only measured rates (initial-rate tables or log–log graphs) tell you the rate law, and the rate law in turn tests which mechanisms are possible.
Summary
Consider the gas-phase synthesis reaction $\mathrm{X}\,(g) + 2\,\mathrm{Y}\,(g) \rightarrow \mathrm{Z}\,(g)$, whose empirical rate equation is…
Consider the gas-phase synthesis reaction $\mathrm{X}\,(g) + 2\,\mathrm{Y}\,(g) \rightarrow \mathrm{Z}\,(g)$, whose empirical rate equation is $v = k\,[\mathrm{X}]\,[\mathrm{Y}]^2$. In a fixed-volume reactor $V_1$ at temperature $T$, the reaction proceeds with an initial rate $v_1$. a) Explain quantitatively how the initial rate changes if, keeping the temperature and volume constant, the amount of $\mathrm{X}$ is tripled and the amount of $\mathrm{Y}$ is halved. b) If the initial mixture is transferred to a vessel of volume $V_2 = V_1 / 3$ at the same temperature, calculate the factor by which the reaction rate is multiplied relative to the original rate in $V_1$.
The gas-phase thermal decomposition reaction $2\,\mathrm{N_2O_5}\,(g) \rightarrow 4\,\mathrm{NO_2}\,(g) + \mathrm{O_2}\,(g)$ at $318\,\mathrm{K}$ is analysed. The plot…
The gas-phase thermal decomposition reaction $2\,\mathrm{N_2O_5}\,(g) \rightarrow 4\,\mathrm{NO_2}\,(g) + \mathrm{O_2}\,(g)$ at $318\,\mathrm{K}$ is analysed. The plot of $\ln v_0$ against $\ln [\mathrm{N_2O_5}]_0$ for four independent experiments is observed to give a straight line of gradient $1.00 \pm 0.02$ with a y-intercept equal to $-6,215$. a) Determine, with justification, the order of reaction and the value of the rate constant $k$ with its corresponding units. b) Consider a proposed mechanism consisting of three elementary steps:
1) $\mathrm{N_2O_5} \rightleftharpoons \mathrm{NO_2} + \mathrm{NO_3}$ (fast equilibrium, constants $k_1$ and $k_{-1}$)
2) $\mathrm{NO_2} + \mathrm{NO_3} \rightarrow \mathrm{NO_2} + \mathrm{O_2} + \mathrm{NO}$ (slow step, constant $k_2$)
3) $\mathrm{NO} + \mathrm{NO_3} \rightarrow 2\,\mathrm{NO_2}$ (fast step, constant $k_3$)
Deduce analytically the expression for the rate law predicted by this mechanism and check whether it reproduces the experimental order. The logarithms are applied to the numerical values of the rates in $\mathrm{mol}\times\mathrm{L}^{-1}\times\mathrm{s}^{-1}$ and of the concentrations in $\mathrm{mol}\times\mathrm{L}^{-1}$, taking those units as references.