At \(800 \mathrm{K},\) the equilibrium constant for the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{A}(g)\) is \(K_{c}=3.1 \times 10^{-4}\) . (a) Assuming both forward and reverse reactions are elementary reactions, which rate constant do you expect to be larger, \(k_{f}\) or \(k_{r}\) ? (b) If the value of \(k_{f}=0.27 \mathrm{s}^{-1},\) what is the value of \(k_{r}\) at 800 \(\mathrm{K} ?(\mathrm{c})\) Based on the nature of the reaction, do you expect the forward reaction to be endothermic or exothermic? (d) If the temperature is raised to \(1000 \mathrm{K},\) will the reverse rate constant \(k_{r}\) increase or decrease? Will the change in \(k_{r}\) be larger or smaller than the change in \(k_{f} ?\)

Step by step solution

01

Equilibrium Constant Relationship

Equilibrium constants can be related to the forward and reverse reaction rate constants with the following formula: \[ K_c = \frac{k_f}{k_r} \] Here, Kc is the equilibrium constant, kf is the forward rate constant and kr is the reverse rate constant. Step 2: Determine the larger rate constant

02

Comparing kf and kr

To determine which rate constant is larger, we can analyze their respective values when the reaction is at equilibrium. Since Kc is 3.1 x 10^(-4) and this value is less than 1, it indicates that the denominator (kr) is larger than the numerator (kf). Thus, the reverse rate constant, kr, is larger than the forward rate constant, kf. Step 3: Calculate the reverse rate constant (kr)

03

Calculating kr

Given the value of the forward rate constant, kf = 0.27 s^(-1), we can find the reverse rate constant, kr, using the formula for the equilibrium constant as seen in Step 1. \[ kr = \frac{k_f}{K_c} \] \[ kr = \frac{0.27 \ \mathrm{s}^{-1}}{3.1 \times 10^{-4}} \] Now, calculate the value of kr: \[ kr \approx 870.97 \ \mathrm{s}^{-1} \] Step 4: Determine the endothermic or exothermic nature of the forward reaction

04

Endothermic or Exothermic

Since the value of Kc increases with temperature (given in part d of the problem), it indicates that the forward reaction is endothermic. The equilibrium shifts for an endothermic reaction to favor the products when the temperature is increased. Step 5: Analyzing how rate constants change with temperature

05

Changes in Rate Constants with Temperature

If the temperature is raised to 1000 K: (a) The reverse rate constant kr will decrease because the equilibrium shifts to favor the products (due to the endothermic nature of the forward reaction). (b) The change in kr will be smaller than the change in kf. The reaction becomes more product-favored at higher temperatures, which indicates that the forward reaction rate will increase by a larger amount compared to the reduction in the reverse reaction rate.

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