# Understanding Free Energy, Reaction Quotients, and Spontaneity in Chemical Reactions

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## Introduction

Understanding the relationship between free energy and the reaction quotient is crucial in comprehending how chemical reactions proceed. This article demystifies the concept of free energy, particularly the change in free energy (
(\Delta G)), and its relationship with the reaction quotient (
(Q)), explaining their significance in evaluating the spontaneity of reactions.

\n## The Foundations of Free Energy and Reaction Quotient

### What is Free Energy?

Free energy is a thermodynamic potential that measures the capacity of a system to do work. The change in free energy ( (\Delta G)) is critical in predicting whether a reaction will occur spontaneously or not.

**(\Delta G < 0)**: Reaction is spontaneous in the forward direction.**(\Delta G > 0)**: Reaction is nonspontaneous in the forward direction.**(\Delta G = 0)**: Reaction is at equilibrium.

### What is Reaction Quotient (

(Q))?

The reaction quotient (
(Q)) represents the ratio of the concentrations (or partial pressures) of products to reactants at any given point during the reaction, similar in form to the equilibrium constant (
(K)).

### The Equations

To understand the connection between these concepts, we use the equation:

[ \Delta G = \Delta G^0 + RT \ln Q ]

Here:

- (\Delta G^0): Standard change in free energy.
- (R): Gas constant.
- (T): Temperature in Kelvin.

## Example: Synthesizing Ammonia

### Given Conditions

Consider the synthesis of ammonia from nitrogen gas and hydrogen gas:

[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) ]

**Partial Pressures at 25°C**:

- N2: 1 atm
- H2: 1 atm
- NH3: 1 atm

The standard change in free energy, (\Delta G^0), is -33.0 kJ.

### Calculating Reaction Quotient

To find the reaction quotient,

[ Q_p = \frac{(P_{\text{NH}*3})^2}{(P*{\text{N}*2})(P*{\text{H}_2})^3} ]

Since all gases are at 1 atm:

[ Q_p = \frac{1^2}{1 \times 1^3} = 1 ]

### Finding (\Delta G)

Plugging values back into our equation:

[ \Delta G = -33.0 imes 10^3 J + (8.314 imes 298) \ln(1) ]

Since (\ln(1) = 0),

[ \Delta G = -33.0 imes 10^3 J = -33.0 ext{ kJ} ]

This indicates that the reaction is spontaneous, and more ammonia will be produced.

## New Conditions: Partial Pressures of 4.0 atm

### The Reaction Quotient

If all partial pressures are now at 4.0 atm,

[ Q_P = \frac{(4.0)^2}{(4.0)(4.0)^3} = \frac{16}{64} = 0.25 ]

### Finding (\Delta G) Again

Using the new (Q_P):

[ \Delta G = -33.0 imes 10^3 J + (8.314 \times 298) \ln(0.25) ]

Calculating:

[ \Delta G = -33.0 imes 10^3 + (8.314 \times 298)(-1.386) ]

Let’s find the result:

[ \Delta G \approx -39.9 ext{ kJ} ]

### Understanding the Changes

With a negative (\Delta G), the reaction is still spontaneous, moving more toward producing ammonia, indicating the system has too many reactants. As the reaction progresses, the numerator (ammonia) increases, while the denominator (nitrogen and hydrogen) decreases, increasing (Q).

\n## Changes as (Q) Increases

As we further manipulate (Q) to equal 100, we calculate:

[ \Delta G = -33.0 imes 10^3 + (8.314 \times 298) \ln(100) ]

Calculating gives us:

[ \Delta G \approx -21.6 ext{ kJ} ]

Though (\Delta G) is less negative, indicating the reaction is still spontaneous, we are approaching equilibrium.

## At Equilibrium

When (Q = K), we expect:

[ \Delta G = 0 ]

### Verifying Equilibrium

Substituting (K = 6.1 imes 10^5):

[ \Delta G = -33.0 imes 10^3 + R(298)\ln(6.1 imes 10^5) ]

This calculation indeed leads us to zero. Therefore, at equilibrium, no further spontaneous reaction takes place; the system balances itself with equal free energy between products and reactants.

## Conclusion

Understanding the dynamics between free energy, reaction quotient, and spontaneity is vital in thermodynamics and chemical kinetics. By analyzing how changes in (Q) influence (\Delta G), we can grasp how reactions progress towards equilibrium, emphasizing the natural tendencies of chemical processes to achieve balance. With real-world applications in chemical production, maintaining this equilibrium is essential for maximizing yield and efficiency in industrial processes.

- [Voiceover] The relationship between the change in free energy and Q, the reaction quotient, is very important to understand.

The change in free energy or delta G is the instantaneous difference in free energy between the reactants and the products.

Q is our reaction quotient; It tells us where we are in the reaction, and remember, it has the same form

as the equilibrium constant K. Delta G zero is the standard change in free energy, so the change in free energy under standard conditions.

R is the gas constant and T is the temperature in Kelvin. Remember when delta G is less than zero, so when delta G is negative,

the reaction is spontaneous in the forward direction. When delta G is greater than zero so when delta G is positive,

the reaction is nonspontaneous in the forward direction. And when delta G is equal to zero the reaction is at equilibrium.

Let's apply this equation to an example. Here we're trying to synthesize ammonia so our goal is to calculate the change in free energy.

So we're trying to find delta G for the following reaction at 25 degrees C given the following set of partial pressures.

So nitrogen gas plus hydrogren gas gives us ammonia gas, and at 25 degrees C, the standard change in free energy, delta G zero is equal to negative 33.0 kilojoules.

And remember, for gasses it's easier to measure pressures than it is to measure concentrations so at this instant in time the partial pressures

for all these gasses are one atmosphere, so one atmosphere for nitrogen, one atmosphere for hydrogen,

We're trying to find delta G so let's write down our equation. Delta G is equal to delta G zero

plus RT natural log of Q. So we need to find Q, we need to find our reaction quotient. Remember, it has the same form as the equilibrium constant.

For our equilibrium constant we had the concentration of products over the concentration of reactants, and we raised the concentration

to the power of the coefficient, but here we're dealing with partial pressures instead of concentrations.

So I'm gonna right Q sub P here to remind ourselves that we're working with partial pressures. So that would be the partial pressure of our product,

so the partial pressure of ammonia raised to the power of the coefficient, so raised to the second power,

over the partial pressure of nitrogen, so the partial pressure of nitrogen, raised to the power of the coefficient,

and our balanced equation here, the coeffecient is a one so this is raised to the first power times the partial pressure hydrogen

raised to the power of the coefficient, which is three, so raised to the third power. So everything is at one atmosphere at this moment in time.

And all that, of course, is equal to one. So at this moment in time our reaction quotient Q is equal to one.

So to solve for the change in free energy, to find delta G, we take Q and we plug it into our equation. What is the natural log of one?

That of course is equal to zero. So this is equal to zero and we find that the change in free energy, delta G,

is equal to the standard change in free energy delta G zero, which, at this temperature, is equal to negative 33.0 kilojoules.

So the change in free energy, delta G, is equal to negative 33.0 kilojoules. Let's talk about a few things here.

First, we got a negative value for delta G. We got negative 33.0, so our reaction is spontaneous in the forward direction.

So we're gonna make more of our products. We're gonna make more ammonia here since delta G is negative.

It's our driving force to make more of our products. Notice that delta G, in this example, delta G is equal to delta G zero,

and that makes perfect sense because delta G zero is the standard change in free energy, it's the change in free energy under standard conditions,

which is what we have here, we're under standard conditions, everything is at one atmosphere, so that makes sense. And finally you could write, if you wanted to,

you could write kilojoules per mole of reaction, indicating for this reaction, how it's written right here, that is the change in free energy.

So when one mole of nitrogen combines with three moles of hydrogen to give us two moles of ammonia, delta G, the change in free energy for that reaction,

is negative 33.0 kilojoules. Now let's get a new set of partial pressures. Now let's say all the gasses have

a partial pressure of 4.0 atmospheres, and when all the gasses have that partial pressure what is the change in free energy at that moment

for our reaction at 25 degrees C? So we're trying to find the change in free energy, we're trying to find delta G,

and we know delta G is equal to delta G zero plus RT natural log of Q. So we need to find Q, the reaction quotient.

This is QP. And from the previous example we know how to find Q, so I'll go through this a little bit faster.

This would be 4.0 to the second power. So 4.0 squared, over 4.0 to the first power, so 4.0 to the first times 4.0 to the third.

So 4.0 to the third. 4.0 squared is 16, over four times-- Let's see. Four cubed is 64, so what is Q equal to?

So at that moment the reaction quotient is equal to .0625, at the moment when all of our partial pressures are four atmospheres.

So this is our reaction quotient. We're gonna plug this into our equation and we're gonna solve for delta G.

So delta G is equal to delta G zero, the standard change in free energy. At 25 degrees C delta G zero is negative 33 kilojoules.

So we have negative 33.0 kilojoules. We need to make that joules so I'm gonna say times 10 to the third joules.

Plus R is the gas constant, so that's 8.314. 8.314, and that's joules over mole times K.

Since the gas constant is in joules per mole we need to make the delta G zero in joules per mole too. Next we have the temperature.

The temperature of our reaction is 25 degrees C, that needs to be in Kelvin. So 25 degrees C is 298 Kelvin.

So we write in here 298 Kelvin, so Kelvin would cancel out here. Then we're going to multiply that by the natural log of Q.

Let's start with the natural log of .0625. The natural log of .0625 is equal to that value, which we need to multiply by 298

times 10 to the third. And that should give us our change in free energy, which is equal to negative 39.9 kilojoules per mole,

It's just easier to think in kilojoules. And remember this is moles of our reaction, so for this specific reaction where one mole of nitrogen

combines with three moles of hydrogen to give us two moles of ammonia. So how it's written, delta G for the reaction

how it's written is equal to negative 39.9 kilojoules. Delta G is negative so we know our reaction is spontaneous in the forward direction.

The reaction is spontaneous in the forward direction which means we're going to make more of our products. Let's think about how far away we are from equilibrium.

Q, our reaction quotient, is .0625. For this reaction at 25 degrees C the equilibrium constant, which would be KP,

is equal to 6.1 times 10 to the fifth. Our value for the reaction quotient was .0625. That's much, much smaller.

That's much smaller than the equilibrium constant. So Q is much smaller than K. So we know we have too many reactants and too few products.

We have a driving force, delta G is negative. We have a driving force to make more of our products. So the reaction moves forward to make more products here.

So we know that, by just looking at the sign for delta G, which we know is spontaneous. We also know that from comparing

the reaction quotient Q to K. So because Q is not equal to K we're not at equilibrium here.

We are away from equilibrium and we have a spontaneous reaction. So we're gonna make more of our products.

The reaction proceeds to the right, we make of-- overall, I should say. We make more of our products

and we lose some of our reactants. What happens to Q? As the reaction progresses we make more of our products

so the numerator should increase, and we lose some of our reactants so the reactants should decrease.

And so you can think about that, an increase in the numerator, a decrease in the denominator, that means that as the reaction progresses,

as the reaction moves to the right to make more products, Q should increase. So as Q increases what happens to delta G?

Let's think about that next. If we get an increase in Q what happens to the change in free energy?

Let's just make up a number here. Let's say that Q is now equal to 100. So we're at a different moment in time.

Here Q was equal to .0625. Let's increase that to Q is equal to 100. So if we increase Q what happens to delta G?

Let's just plug the numbers into the equation and let's see for ourselves. Delta G is equal to negative 33.0 times 10 to the third.

That's joules per mole. Actually, let's just leave out units here to give us a little bit more room.

That's delta G zero, plus 8.314, our gas constant, times our temperature, we're still at 298K. So the only difference now is

we're substituting 100 in for Q. So we've increased the value for the Q. What happens to delta G? Let's do the math.

We need to find the natural log of 100 and we multiply that by 298 and we multiply that by 8.314.

So we get, in kilojoules per mole that would be negative 21.6 kilojoules per mole. So delta G is equal to negative 21.6

kilojoules per mole of reaction. So we increased Q, we increased Q, what happened to delta G? We went from a delta G of negative 39.9

to a delta G of negative 21.6. So we're getting closer to zero as the reaction proceeds to the right.

We're still spontaneous, we still have a negative value for delta G, so we still have a driving force to make more products.

Q is still less than K. Q is 100 and K is 6.1 times 10 to the fifth. So we're still gonna go to the right and make more products,

our reaction is still spontaneous. What happens at equilibrium? What happens at equilibrium? We know already

that delta G should be equal to zero at equilibrium. So equilibrium, delta G should be equal to zero. And Q should be equal to K.

The reaction quotient is equal to the equilibrium constant. Let's plug that into our equation and see if that's true. We have delta G is equal to delta G zero,

negative 33.0 times 10 to the third plus the gas constant is 8.314. Our temperature is still 298.

But instead of writing Q we're gonna plug in K, we're gonna plug in the equilibrium constant, which I already gave to you up here.

It's 6.1 times 10 to the fifth. So now we're doing the natural log of K. We plug in the natural log of K

which is 6.1 times 10 to the fifth. And let's see what happens to delta G. Now we're going to find the natural log

of 6.1 times 10 to the fifth. And we need to multiply that by 298 and then multiply that by 8.314.

So we get positive 33 kilojoules. So positive 33 kilojoules, or 33.0 times 10 to the third. So notice what happens.

Over here we have delta G zero is negative 33.0 times 10 to the third, and all of this on the right gives us

positive 33.0 times 10 to the third. So that of course is equal to zero. So the change in free energy is now zero.

And we are at equilibrium. Q is equal to K, we plugged into our equation and we found that delta G is indeed equal to zero so

there's no more driving force to make more of our products. There's no more driving force to make more of our products because we are at equilibrium.