Kinetics 1: Rate Laws and Constants

The rate of reaction is the change in concentration of a reactant or product per unit time.

The rate expression or rate law is an equation that relates the rate of reaction to the concentrations of the reactants.

The rate constant is a proportionality constant that relates the rate of reaction to the concentrations of the reactants.

<p>PA = Pt - PB.</p> Signup and view all the answers

<p>The data are indeed consistent with a second-order reaction. For a second-order reaction, the rate law is given by: rate = k[A]², where k is the rate constant and [A] is the concentration of reactant A. As the reaction proceeds, the concentration of A decreases, but the rate of reaction decreases more rapidly. This is because the rate is proportional to the square of the concentration of A. In this case, the rate constant is k = 8.06 x 10^6 Torr⁻¹s⁻¹.</p> Signup and view all the answers

<p>To determine the rate constant, we can use any two data points from the table and the integrated rate law for a second-order reaction. For instance, using the data at t=0 and t=200s, we have: 1/[A]t - 1/[A]0 = kt Substituting the values for [A]t, [A]0, and t, we obtain k = 8.06 x 10⁶ Torr⁻¹s⁻¹</p> Signup and view all the answers

<p>The activation energy can be calculated using the Arrhenius equation: k = Ae^(-Ea/RT), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. Taking the natural logarithm of both sides, we get: ln(k) = ln(A) - Ea/(RT), The activation energy can be obtained by Ea = - R(ln(k₂)- ln(k₁)) / (1/T₂ - 1/T₁) where k1, k2 are the rate constants at temperatures T1 and T2.</p> Signup and view all the answers

<p>We can apply the steady-state approximation to the reaction mechanism, which assumes that the concentration of the intermediate O is constant over time. This allows us to set the rate of formation of O equal to the rate of its consumption: k1[O3] = k2[O3][O] + k3[O2][M][O] Solving for [O], we get: [O] = k1[O3] / k2[O3] + k3[O2][M].</p> Signup and view all the answers

<p>Species M represents a third body, typically an inert gas molecule, that collides with the reactive intermediate (O). This collision helps to remove excess energy from the system and stabilizes the newly formed ozone molecule, preventing it from immediately decomposing back into O2 and O. This is why M is included on both sides of the reaction equation, as it plays a crucial role in the overall rate of ozone formation.</p> Signup and view all the answers

<p>The rate of disappearance of ozone is given by: d[O3]/dt = -k1[O3] + k3[O2][M][O] Substituting the expression for [O] that was derived earlier, we get: d[O3]/dt = -2k1k2[O3]²/ k2[O3] + k3[O2][M]</p> Signup and view all the answers

<p>The steady-state approximation assumes that the rate of formation of the intermediate O is equal to the rate of its consumption. This assumption is valid if the concentration of the intermediate is much lower than the concentrations of the reactants and products. In this case, the assumptions are justified if the rate constant k2 for the second reaction is much smaller than the rate constant k3 for the third reaction. This ensures that the intermediate O is rapidly consumed, resulting in a relatively constant concentration.</p> Signup and view all the answers

<p>A rate law in chemical kinetics is a mathematical expression that describes the rate of a reaction as a function of the concentrations of the reactants. For example, the rate law for the reaction 2A + B → C could be written as rate = k[A]²[B], where k is the rate constant. The overall order of the reaction is the sum of the exponents of the concentration terms in the rate law. In the example above, the overall order would be 3. The concept of overall order is not applicable to all rate laws. This applies to any rate law for reactions that are not elementary steps (reactions that occur in a single step without any intermediates). Elementary reactions are those where the rate law is consistent with the stoichiometry of the reaction. For example, the reaction A + B → C would have a rate law of rate = k[A][B] as the order of each reactant corresponds to its coefficient in the balanced equation.</p> Signup and view all the answers

<p>To derive the rate law, we need to apply the steady-state approximation to the intermediate CuH+. Assuming the concentration of CuH+ remains constant over time, we can equate the rate of formation to the rate of consumption, resulting in: k1[Cu2+][H2] = k-1[CuH+][H+] + k2[Cu2+][CuH+] Solving for [CuH+] we get: [CuH+] = k1[Cu2+][H2] / k-1[H+] + k2[Cu2+] Then, we can use the rate of the slowest step (the rate-determining step) which is the rate of the reaction: rate = k2[CuH+][Cu2+] Substituting the expression for [CuH+], we obtain the rate law: rate = k1k2[Cu2+]²[H2] / k-1[H+] + k2[Cu2+]</p> Signup and view all the answers

<p>To demonstrate that the provided rates align with the mechanism, we can substitute the experimental values into the theoretical rate law derived earlier and evaluate whether it accurately predicts the observed rates. Using the given data, we can compute k1 and k-1/k2 using any two sets of data points and the rate law derived earlier: rate = k1k2[Cu2+]²[H2] / k-1[H+] + k2[Cu2+] Substituting the values for the concentrations of Cu2+, H2, and H+, we can then solve for the rate constant k1 and the ratio k-1/k2. k1 = 1.73 x 10-5 mol dm-3 s-1, and k-1/k2 =0.0037 mol dm-3 The consistency of the calculated rate constants across different data points validates this mechanism. Furthermore, the values of k1 and k-1/k2 fit well within the range of typical rate constants observed in chemical reactions, further reinforcing the validity of the proposed mechanism.</p> Signup and view all the answers

<p>The rate of conversion of A into B is given by the rate of the slowest step in the mechanism, which is step 2. Step 2 is a unimolecular reaction involving the decomposition of A* into B with a rate constant k2. The rate of conversion of A to B is given by: d[B]/dt= k2[A*] We can express [A*] in terms of [A] using the steady-state approximation for A*: Assuming a steady-state condition for A*, we can set its rate of formation equal to its rate of consumption: k1[A]² = k-1[A][A*] + k2[A*] Solving for [A*], we get: [A*] = k1[A]² / k-1[A] + k2 Substituting this expression for [A*] in the rate law, we obtain: d[B]/dt = k1k2[A]² / k-1[A] + k2</p> Signup and view all the answers

<p>The rate of the reaction is determined by the rate of the slowest step, which is the unimolecular decomposition of A* into B (k2). At high pressure, the rate of formation of A* (k1) is much faster than the rate of its decomposition (k2). The unimolecular decomposition of A* will be the rate-determining step and the overall reaction will be first-order. As we decrease the pressure, the rate of formation of A* decreases, and therefore A* becomes less stable and decomposes more rapidly. This effect is more pronounced at low pressure where the [A] is low. As a result, the unimolecular decomposition of A* will no longer be the rate-determining step and the reaction will become second order. Inert gases have a significant effect on unimolecular reactions because they reduce the rate of A* decomposition by increasing collisions and reducing the lifetime of A*. In this case, an inert gas has the effect of stabilizing A* because it is more likely to collide with A* and transfer energy to it. This process allows for a greater proportion of A* to decompose into B and increases the rate constant for this reaction.</p> Signup and view all the answers

<p>The reaction scheme for the cis-trans isomerisation of an alkene involves the formation of an activated complex, which can either decompose back to the original isomer or convert to the other isomer. Therefore, the rate of the isomerisation reaction depends on the rate of formation of the activated complex. The reaction scheme is: A + M → A* + M, k1 A* → A, k-1 A* → B, k2 where A is the alkene molecule, M is a third body, A* is the activated complex, and B is the other isomer. The rate of formation of the activated complex (A*) is proportional to the concentration of alkene. So, as the concentration of alkene increases, the rate of formation of A* also increases. This results in an increase in the rate of the isomerisation reaction. The rate constant k1 for the formation of the activated complex increases with the collision number, so the rate of isomerisation also increases with the concentration of the alkene. The rate constant k2 for the isomerisation step is independent of the concentration of the alkene. However, at high concentrations of alkene, the rate constant k-1 for the decomposition of the activated complex will become significant, the rate of the overall reaction will decrease, and the reaction will approach second order. Thus, at high concentrations of the alkene, the reaction is closer to second order kinetics. This is consistent with the provided data because as the concentration of the alkene increases, the rate constant increases but does not increase linearly, which is consistent with the reaction approaching second order kinetics.</p> Signup and view all the answers

<p>At high pressure, the rate of the isomerisation reaction is limited by the rate of formation of the activated complex A*, so the rate constant k₁ is a good approximation of the limiting value of the effective first-order rate constant. k1 = 1.8 x 10^5 s-1 .</p> Signup and view all the answers

<p>The assumption that only unimolecular reactions follow the scheme shown is not necessarily valid. In cases where the reaction involves a combination of unimolecular and bimolecular steps, the reaction kinetics can become complex and less straightforward, deviating from the simple predictions of the scheme. Another factor is that the mechanism only considers the energy transferred in a single collision. In reality, multiple collisions are possible, and the energy transfer between collisions also plays a role. These factors can make the mechanism more complex and can lead to deviations from the predictions of a single-collision model. Additionally, collisions with the third body can cause energy transfer. When the concentration of the reactant molecules decreases, the frequency of collision diminishes, and the system deviates from the unimolecular scheme. The rate of the reaction is likely to decrease at lower concentrations, which can deviate from the predictions of this simple scheme.</p> Signup and view all the answers

<p>The half-life of a chemical reaction is the time taken for the concentration of a reactant to decrease to half of its initial value. The reaction is a bimolecular reaction as it involves two reacting species. The rate equation for this reaction is: d[OH]/dt = - k1[OH][C2H6] At the initial time, the concentration of [OH]=[C2H6]= a0, so the rate equation becomes d[OH]/dt = - k1a0[OH] We can integrate this equation to obtain the time taken for [OH] to decrease to half of its initial value: [OH] = a0 exp(-k1 a0 t) When t = t½, [OH] = a0/2 Substituting these values into the equation above, we get: t½ = (a0k1)⁻¹ Therefore, the half-life of the reaction is given by (a0k1)⁻¹.</p> Signup and view all the answers

<p>The half-life of a unimolecular reaction is independent of the initial concentration of the reactant. Therefore, the OH radical half-life will remain the same even though the initial concentration of C2H6 is significantly higher. So, the half-life of the OH radical will remain 44s.</p> Signup and view all the answers

<p>From the table, we can see that the half-life of the OH radical decreases with increasing temperature. This indicates that the rate of the reaction increases with increasing temperature, which is consistent with the Arrhenius equation. As temperature increases, the molecules have more energy and are more likely to collide with enough energy to overcome the activation energy barrier and react. Furthermore, the data show that the decrease in half-life as we increase the temperature is not linear. This is because the Arrhenius equation is an exponential equation, and the rate constant is exponentially dependent on temperature. We can also calculate the activation energy from these data. For example, using the data for 300 K and 450 K, the activation energy can be calculated using the Arrhenius equation: ln(k1/k2) = -Ea/R (1/T1 - 1/T2) Assuming the reaction is first order, we can use the half-life values to get the rate constant at each temperature. From this, we can conclude that, for this reaction, the half-life of the OH radical is inversely proportional to temperature, which is consistent with the Arrhenius equation.</p> Signup and view all the answers

<p>The recombination reaction of oxygen atoms is an example of a termolecular reaction, involving three reactant species. In this case, the third body (M) plays a crucial role in stabilizing the newly formed O2 molecule. The reaction proceeds in two steps: O + O → O2* A collision between O molecules is required to produce an unstable O2* molecule. O2* + M → O2 The unstable O2* molecule needs a third body (M) to remove excess energy, so it can be stabilized into a stable O2 molecule. At a higher temperature, the average kinetic energy of the system is higher, leading to the following: 1. The rate of collision between O atoms increases, resulting in a higher frequency of forming unstable O2* molecules. 2. The rate of collision between O2* and M decreases due to the faster relative speeds between molecules. Therefore, with a higher temperature, the unstable O2* molecule is more likely to decompose back into two O atoms rather than being stabilized into O2 by a third body. As a result, the rate of the overall recombination reaction decreases with increasing temperature, leading to an increase in the half-life of oxygen atoms.</p> Signup and view all the answers