Consider the system A(g) ⟶B(g) at 25^∘ C. a. Assuming that GA^∘=8996 J / mol and GB^∘=11,718 J

Consider the system A(g) ⟶B(g) at 25^∘ C. a. Assuming that...

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at $25^{\circ} \mathrm{C}$. a. Assuming that $G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}$ and $G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}$, cal- culate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if $1.00 \mathrm{~mol} \mathrm{~A}(\mathrm{~g})$ at $1.00$ atm and $1.00 \mathrm{~mol} \mathrm{~B}(g)$ at $1.00 \mathrm{~atm}$ are mixed at $25^{\circ} \mathrm{C}$. c. Show by calculations that $\Delta G=0$ at equilibrium.

Consider the system
$$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$
at $25^{\circ} \mathrm{C}$.
a. Assuming that $G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}$ and $G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}$, cal-
culate the value of the equilibrium constant for this reaction.
b. Calculate the equilibrium pressures that result if $1.00 \mathrm{~mol} \mathrm{~A}(\mathrm{~g})$ at $1.00$ atm and $1.00 \mathrm{~mol} \mathrm{~B}(g)$ at $1.00 \mathrm{~atm}$ are mixed at $25^{\circ} \mathrm{C}$.
c. Show by calculations that $\Delta G=0$ at equilibrium.

Key Concepts

Standard Gibbs Free Energy Change

This concept refers to the change in Gibbs free energy (?G°) under standard conditions for a reaction. It indicates whether a reaction is spontaneous or non-spontaneous and provides a quantitative measure of the driving force of a chemical reaction. A negative ?G° means that the reaction proceeds spontaneously under standard conditions, while a positive ?G° indicates non-spontaneity.

Equilibrium Constant

The equilibrium constant (K) quantifies the ratio of product concentrations to reactant concentrations at equilibrium. Its relationship with the standard Gibbs free energy change is given by the equation ?G° = -RT ln K, linking thermodynamics with chemical equilibrium. This relation allows one to calculate K from ?G° and helps predict the extent of a reaction under given conditions.

Reaction Quotient and Equilibrium

The reaction quotient (Q) is similar to the equilibrium constant but applies to any moment in time, not just at equilibrium. At equilibrium, Q equals K, and there is no net change in the concentrations or pressures of the reactants and products. The concept of Q is essential for understanding how a system approaches and achieves equilibrium.

Equilibrium Condition (?G = 0)

At equilibrium, the Gibbs free energy change (?G) for the reaction is zero. This means that the system has reached a state where there is no net driving force for the reaction to proceed in either the forward or reverse direction. This concept is fundamental in thermodynamics as it defines the point of balance between the forward and reverse processes.

Transcript

00:02 In this problem, we're asked to consider the reaction that's given right here, and we are also given a couple of other numbers, and we are asked to find first our equilibrium constant.

00:14 Okay, so let's find our equilibrium constant.

00:20 First, we're going to have to find our delta g.

00:28 Our delta g for our reaction will be equal to b minus a, and this will be 11 ,718 minus 8 ,996 is 27 ,22.

00:55 So now i'm going to use my k will be equal to e to the negative of my delta g divided by rt.

01:16 K, and that's it, will be equal to, i always make these crappy e's, negative 2722 j per mole divided by 8 .314 j per mole k times, i believe we're at 25 degrees c.

01:55 Okay, let's do this.

02:06 Second e, negative 2722 divided by 8 .314 divided by 298 is 0 .33 is my k.

02:27 I'm pretty happy with that.

02:29 Let's go 3, 3, 3, 3.

02:43 Okay.

02:43 B, was there a b? i don't even remember if there's a b.

02:52 What is the equilibrium pressures pressures if 1 .00 moles of a at 1 .00 atm and 1 .00 moles of b at 1 .00 atm are mixed at 25 degrees c.

03:35 Okay, i keep looking at a particular problem here and i'm having trouble.

03:47 I'm going to pause for a sec because i'm having trouble with another program.

03:51 I'll go fix this later.

03:53 Okay, so my equilibrium pressure.

03:57 Let's get started on this one...