# Understanding Delta-G and Equilibrium Constants in Chemical Reactions

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### Introduction

In the field of chemistry, understanding the dynamics of reactions is crucial. One of the key concepts that govern reaction behavior is the change in free energy, denoted as delta-G (ΔG). In this article, we will dive into the relationship between ΔG and the equilibrium constant (K). We will explore how these concepts inform us about the spontaneity of reactions and the composition of chemical mixtures at equilibrium.

### The Fundamentals of Delta-G

Delta-G, or the change in free energy, is a thermodynamic parameter that indicates the spontaneity of a reaction. A reaction is considered spontaneous if the ΔG is negative, indicating that the reaction can proceed without external energy input. Conversely:

**ΔG < 0**: Reaction is spontaneous**ΔG = 0**: Reaction is at equilibrium**ΔG > 0**: Reaction is non-spontaneous

### Reaction Quotient and Equilibrium Constant

In our previous discussion, we explored how the reaction quotient (Q) interacts with ΔG. At equilibrium, this quotient equals the equilibrium constant (K). When plugging values into the evaluated equations, we notice:

- If ΔG = 0, then Q = K

This illustrates that at equilibrium, there is no net change in the concentrations of reactants and products.

### Equations Relating ΔG and K

To link ΔG with K mathematically, we have:

[ \Delta G^0 = -RT \ln(K) ]

Where:

**ΔG<sup>0</sup>**is the standard change in free energy**R**is the gas constant (8.314 J/(mol·K))**T**is the absolute temperature in Kelvin**K**is the equilibrium constant

This equation tells us that at equilibrium, when ΔG is zero, the equation transforms to: [ K = e^{(\Delta G^0)/(RT)} ]

### Analyzing Specific Reactions

Let's evaluate how ΔG and K behave under different conditions by analyzing specific reactions at varying temperatures.

#### Example 1: Synthesis of Ammonia at 298 K

In studying the synthesis of ammonia at **298 K**, we find:

**ΔG<sup>0</sup> = -33.0 kJ**

Plugging this into our equation: [ -33.0 \times 10^3 = -8.314 \times 298 \ln(K) ]- Calculate:
- Divide by the constants
- Find ln(K)
- Solving gives us:
**K = 6.1 × 10⁵**

This indicates that at equilibrium, products are favored over reactants since K is significantly greater than one.

#### Example 2: Reaction at 464 K

Next, consider the reaction at **464 K**, where:

**ΔG<sup>0</sup> = 0**

Using the equation: [ 0 = -RT \ln(K) ]

Here, the natural log of K equals zero, hence: [ K = e^{0} = 1 ] This result highlights that reactants and products are equally favored in the equilibrium mixture.

#### Example 3: Reaction at 1000 K

Finally, at **1000 K**, where:

**ΔG<sup>0</sup> = +106.5 kJ**

We again use the equation: [ +106.5 \times 10^3 = -8.314 \times 1000 \ln(K) ]

Calculating gives:- find ln(K), which results in
**K = 2.7 × 10⁻⁶**

This time, K being less than one shows that the reactants are favored in the equilibrium mixture.

### Conclusion

In summary, ΔG and the equilibrium constant K are fundamental concepts in understanding chemical reactions. We see that:

- A negative ΔG indicates reaction spontaneity and a strong favoring of products (K >> 1)
- A ΔG of zero represents an equilibrium state where products and reactants coexist equally (K = 1)
- A positive ΔG indicates non-spontaneity with a favoring of reactants (K << 1) By mastering these relationships, chemists can predict the direction and extent of reactions under various conditions, which is vital for theoretical studies and practical applications as well.

- [Voiceover] In the previous video, we looked at the relationship between the change in free energy, delta-G,

and the reaction quotient, Q. And we plugged in different values for Q and we saw how that affected our answer for delta-G.

The sign of delta-G told us if a reaction was spontaneous or not. We also said that at equilibrium,

Q, the reaction quotient is equal to the equilibrium constant, K. And we plugged K into the equation

in free energy is equal to zero. So, there's no difference in free energy between the reactants and the products.

Let's plug in delta-G is equal to zero into our top equation here, so, we have zero is, zero is equal to delta-G zero,

the standard change in free energy, plus R times T, and since we're at equilibrium, delta-G is equal to zero, this would be

the natural log of the equilibrium constant, K. So, we solve for delta-G zero. Delta-G zero is equal to negative RT,

natural log of K. So, we have another very important equation to think about. Delta-G zero is the standard change in free energy,

or the change in free energy under standard conditions. R is the gas constant, T is the temperature in Kelvin, and K is our equilibrium constant.

So, if you're using this equation, you're at equilibrium, delta-G is equal to zero. And we know at equilibrium, our equilibrium constant

tells us something about the equilibrium mixture. Alright, do we have more products or do we have more reactants at equilibrium.

And this equation relates the equilibrium constant K to delta-G zero, the standard change in free energy.

So, delta-G zero becomes a guide to the ratio of the amount of products to reactants at equilibrium, because it's related

to the equilibrium constant K in this equation. If you're trying to find the spontaneity of a reaction, you have to use this equation up here,

and look at the sign for delta-G. So, if you're trying to find if a reaction is spontaneous or not, use this equation.

If you're trying to find or think about the ratio of the amount of products to reactants at equilibrium, then you wanna use this equation down here,

and that ratio is related to the standard change in free energy, delta-G zero. Now we're ready to find some equilibrium constants.

Remember, for a specific temperature, you have one equilibrium constant. So, we're going to find the equilibrium constant

for this reaction at 298 K. So, we're trying to synthesize ammonia here, and at 298 Kelvin, or 25 degrees C,

the standard change in free energy, delta-G zero, is equal to negative 33.0 kilojoules for this balanced equation.

So, let's write down our equation that relates delta-G zero to K. Delta-G zero is equal to negative RT,

natural log of K. Delta-G zero is negative 33.0 kilojoules, so, let's write in here, negative 33.0,

and let's turn that into joules, so times ten to the third, joules. This is equal to the negative,

joules over moles of reaction. So, for this balanced equation, for this reaction, delta-G zero is equal to

negative 33.0 kilojoules. So, we say kilojoules, or joules, over moles of reaction just to make our units work out, here.

Temperature is in Kelvin, so we have 298 K, so, we write 298 K in here, Kelvin would cancel out, and then we have the natural log of K,

our equilibrium constant, which is what we are trying to find. So, let's get out the calculator

and we'll start with the value for delta-G zero which is negative 33.0 times 10 to the third.

So, we're going to divide that by negative 8.314, and we'd also need to divide by 298.

and this is equal to the natural log of the equilibrium constant, K. So, how do we solve for K here?

Well, we would take E to both sides. So, if we take E to the 13.32 on the left, and E to the natural log of K on the right,

this would cancel out and K would be equal to E to the 13.32, so let's do that. So, let's take E

to the 13.32, and that's equal to, this would be 6.1, 6.1 times ten to the one, two, three, four, five.

if you wanted to put in a KP here, you could. So, now we have an equilibrium constant, K, which is much greater than one.

And we got this value from a negative value for delta-G zero. So, let's go back up to here,

and we see that delta-G zero, right, is negative. So, when delta-G zero is less than zero, so when delta-G zero is negative,

what do we get for our equilibrium constant? We get that our equilibrium constant, K, is much greater than one.

So, what does this tell us about our equilibrium mixture? This tells us that at equilibrium, the products are favored over the reactants,

so the equilibrium mixture contains more products than reactants. And we figured that out by using our value for delta-G zero.

Let's do the same problem again, but let's say our reaction is at a different temperature. So now, our reaction is at 464 Kelvin,

so we're still trying to make ammonia here, and our goals is to find the equilibrium constant at this temperature.

At 464 Kelvin, the standard change in free energy, delta-G zero, is equal to zero. So, we write down our equation,

delta-G zero is equal to negative RT, a natural log of the equilibrium constant, K. And this time, for delta-G zero, we're plugging in zero.

So, zero is equal to, we know that R is the gas constant, and we know that the temperature here would be 464 Kelvin.

So, for everything on the right to be equal to zero, the natural log of K must be equal to zero. So, we have zero is equal to the natural log of K.

And now, we're solving for K, we're finding the equilibrium constant. So, we take E to both sides.

So, E to the zero is equal to E to the natural log of K. E to the natural log of K is just equal to K. So, K is equal to E to the zero,

and E to the zero is equal to one. So, when delta-G zero is equal to zero, so let's write this down on here,

so, when your standard change in free energy, delta-G zero, is equal to zero, K is equal to one.

And that means that at equilibrium, your products and your reactants are equally favored. Let's do one more example.

So, let's find the equilibrium constant again at another temperature. So, now we're at 1000 K,

and our standard change in free energy, delta-G zero, is equal to positive 106.5 kilojoules. So, delta-G zero is equal to

negative RT, natural log of K. This time, we're putting in positive 106.5 kilojoules, which is positive 106.5

I'll leave units out of this just to make it a little bit clearer, times the temperature, which is 1000 K,

so this would be 1000 Kelvin times the natural log of the equilibrium constant, K. So, let's do the math there.

So, we're going to take that value and divide it by negative 8.314, and then, we need to divide by 1000,

and that gives us negative 12.81. So, we have negative 12.81 is equal to the natural log of the equilibrium constant.

So, to solve for the equilibrium constant, we take E to both sides, and we get that K is equal to E to the negative 12.81.

So, what is that equal to? E to the negative 12.81 is equal to 2.7 times 10 to the negative six.

So, when delta-G zero is positive, when the standard change in free energy is positive, let's write this one down.

So, when delta-G zero is greater than zero, so, when it's positive, your equilibrium constant, K, is less than one.

Alright, so K is less than one. And we know what that means at equilibrium. The reactants are favored at equilibrium.