The forward and reverse rate constants for the reaction A(g)+B(g) ⇌C(g) are 3.6 ×10^-3 / M ·s an

The forward and reverse rate constants for the reaction...

The forward and reverse rate constants for the reaction $\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$ are $3.6 \times 10^{-3} / M \cdot \mathrm{s}$ and $8.7 \times 10^{-4} \mathrm{~s}^{-1},$ respectively, at $323 \mathrm{~K}$. Calculate the equilibrium pressures of all the species starting at $P_{\mathrm{A}}=1.6 \mathrm{~atm}$ and $P_{\mathrm{B}}=0.44 \mathrm{~atm}$.

The forward and reverse rate constants for the reaction $\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$ are $3.6 \times 10^{-3} / M \cdot \mathrm{s}$
and $8.7 \times 10^{-4} \mathrm{~s}^{-1},$ respectively, at $323 \mathrm{~K}$. Calculate the equilibrium pressures of all the species starting at $P_{\mathrm{A}}=1.6 \mathrm{~atm}$ and $P_{\mathrm{B}}=0.44 \mathrm{~atm}$.

Key Concepts

Forward and Reverse Rate Constants

These are parameters that quantify the speed of the forward and reverse reactions, respectively, under given conditions. Their values determine how quickly a system can approach equilibrium and also relate directly to the equilibrium constant when the ratios of these constants are taken.

Equilibrium Constant

The equilibrium constant is defined as the ratio of the forward reaction rate constant to the reverse reaction rate constant (assuming elementary reaction steps), and it expresses the ratio of the concentrations or pressures of products to reactants at equilibrium. In gas-phase reactions, it can also be expressed in terms of partial pressures.

Law of Mass Action

This principle states that the rate of a chemical reaction is proportional to the product of the concentrations (or partial pressures, in the case of gases) of the reactants, each raised to a power equal to its stoichiometric coefficient. This law is used to derive the expression for the equilibrium constant from the reaction mechanism.

Partial Pressures

In gas-phase reactions, partial pressures function analogously to concentrations in solution-phase reactions. They are used in the calculation of equilibrium when applying the Law of Mass Action, allowing one to relate the measured pressures of gases to the equilibrium position of the reaction.

Transcript

00:01 Alright, so we have our reaction between a and b, making c in the gas phase react...

00:08 It's a gas phase reaction, so all the species are gases, balanced with coefficients of 1.

00:14 To get the equilibrium constant, we can take the ratio of the rate constants...

00:20 I should use lowercase k.

00:27 So let's see, for the forward reaction, it's 3 .6 times 10 to the negative third, and then for the reverse reaction, it's 8 .7 times 10 to the negative fourth.

00:45 Oh, these are in concentration units, so this is actually kc.

00:53 Okay, so let's see, 3 .6 times 10 to the negative third, divided by 8 .7, divided by 10 to the negative fourth, comes out to 4 .138.

01:12 Then to get kp, we can use the relationship kp is equal to kc times rt, raised to the change in the number of moles of gas, which in this case, the change in the number of moles of gas is one gas mole for the products, minus two for the reactants, is negative one.

01:31 So we have 4 .138 times r, which is 0 .0821, times the temperature already given to us in kelvin, 323, raised to the negative one power, comes up to 0 .156.

01:56 All right, now we have a and b making c.

02:03 We know the value of k, and we have some initial conditions.

02:07 Initial pressure of a is 1 .6, and b is 0 .44.

02:12 Nothing is said about c, so we say its pressure is zero, and then we have plus some amount, because it has to go from left to right to make c.

02:21 This is going to be minus and minus...

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