Consider a reaction A2(g)+B2(g) ⇌2 AB(g), with atoms of A shown in red in the diagram and atoms

Consider a reaction A2(g)+B2(g) ⇌2 AB(g), with atoms of A...

Consider a reaction $\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \rightleftharpoons 2 \mathrm{AB}(g),$ with atoms of A shown in red in the diagram and atoms of B shown in blue. (a) If $K_{c}=1,$ which box represents the system at equilibrium? (b) What is the sign of $\Delta G$ for any process in which the contents of a reaction vessel move to equilibrium? (c) Rank the boxes in order of increasing magnitude of $\Delta G$ for the reaction. [Sections 19.5 and 19.7$]$

Key Concepts

Spontaneity and Reaction Direction

When any system not at equilibrium moves towards equilibrium, the process occurs spontaneously, meaning that the change in Gibbs free energy (?G) is negative. This negative value acts as a driving force, indicating that the system is lowering its free energy as it approaches the equilibrium state.

Gibbs Free Energy (?G)

Gibbs free energy is a thermodynamic potential that predicts the direction of a chemical process under constant temperature and pressure. A negative ?G indicates a spontaneous process, while ?G equals zero at equilibrium. This concept links the chemical equilibrium state with the thermodynamic favorability of the reaction.

Reaction Quotient (Q)

The reaction quotient is calculated in the same way as the equilibrium constant but for any given set of concentrations or pressures, not necessarily at equilibrium. Comparing Q to K indicates the direction in which the reaction must proceed to attain equilibrium; when Q is less than K, the reaction will proceed in the forward direction, and when Q is greater than K, it will proceed in the reverse direction.

Equilibrium Constant (K)

The equilibrium constant is a value that quantifies the ratio of the concentrations (or partial pressures) of products to reactants at equilibrium. A K value of 1 indicates that, at equilibrium, the concentrations of reactants and products are roughly equal, reflecting no strong preference for formation of either species.

Chemical Equilibrium

This is the state in which the forward and reverse reactions occur at the same rate so that the concentrations of all reactants and products remain constant over time. The system no longer shows any net change and exists at a minimum of its free energy under the given conditions.

Transcript

00:01 We're taking a look at the equilibrium reaction where a2 and b2 form 2 ab.

00:09 And if we're told what the equilibrium constant is for this particular equilibrium reaction, if we're told, for instance, that the equilibrium constant is one, how would we figure out which of these systems or which combination of these systems is at equilibrium? well, we would have to remember our law of mass action.

00:37 And so when we take a look at this particular reaction, we would have to take a look at our law of mass action.

00:48 So remember that the coefficients become the exponents, the products go in the numerator, and the reactants go in the denominator.

01:01 And we can take a look at the number of particles we have in each case using the law of mass action.

01:08 So for, let me just do this, let me just label them one, two, and three.

01:14 So for one, we can take a look at the number of a -bs where we would have both a red and a blue dot together.

01:26 And then we can take a look at the a -2s where i have a in red.

01:30 So we'd have two red dots together.

01:32 The bs are blue, we would have to have two blue dots together.

01:37 So in box one, what we know is we have three of the a -bs, so it would be three squared, and we have three a -2s and three b -2s, so we can plug those numbers into our law of mass action, and we get in this case one.

01:57 So we know the first box is at equilibrium, but what about the other two? so if we go through this process for the other two, we have one ab in that case, and we have four of the a2s and b2s, which give us a value of 0 .06 for k2.

02:20 If we look at box three, we can do the same thing, and we have seven of the abs.

02:28 Oh, sorry, i can score that one.

02:29 7 abs, and we have one of each of the other ones.

02:34 So we're going to have 49 as our answer here.

02:39 So only box 1 has a k value equal to the equilibrium constant that we want...

Please give Ace some feedback