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### Course: AP®︎/College Chemistry > Unit 9

Lesson 5: Free energy and equilibrium# Nonstandard free energy changes

The standard change in free energy, Δ

*G*°, for a reaction applies only when the reactants and products are in their standard states. In all other circumstances, we must consider the*nonstandard*free energy change, Δ*G*, which can be calculated from Δ*G*° using the equation Δ*G*= Δ*G*° +*RT*ln*Q*. In this equation,*R*is the appropriate gas constant (8.314 J/mol·K),*T*is the temperature in kelvin, and*Q*is the reaction quotient. Created by Jay.## Want to join the conversation?

- I thought the standard gas constant for an ideal gas is 0.08206, as that comes from the ideal gas law?(1 vote)
- The ideal gas constant, R, has a constant value, but it can be expressed using different units and therefore have different units. For example, 1 kilogram and 1000 grams are the same amount of mass, but are different numbers because the units are different.

The constant has a value of 0.082057366080960 (often shortened to 0.08206) when expressed in a unit of ((L∙atm)/(K∙mol)), where volume is in liters, pressure is in atmospheres, temperature is in kelvins, and number of moles is in moles.

Strictly using only SI units, we get a different number. The constant has a value of 8.31446261815324 (often shortened to 8.314) when expressed in a unit of ((m^(3)∙Pa)/((K∙mol)), where volume is in cubic meters, pressure is in pascals, temperature is in kelvins, and number of moles is in moles. Additionally, volume times pressure is equal to work because of pressure-volume work. Work has the SI unit of joules (J), which means that the SI units of volume, cubic meters, and pressure, pascals, when multiplied together yield a joule. So the previous gas constant number, 8.31446261815324, also has the unit J/(K∙mol). We can convert the joule in that unit into a kilojoule (kJ) by dividing the previous number by 1000 since there are 1000 joules in 1 kilojoule. Using a unit of kJ/(K∙mol) yields the number 0.00831446261815324 (shortened to just 0.008314 in the video). So ultimately it’s the same gas constant, it’s just different variations due to unit conversions. We use the different variations of the gas constant based on the units we’re given for a problem. In the video’s problem, Jay wanted the gibbs free energy to be in units of kilojoules so we chose the gas constant which had kJ in its unit.

Hope that helps.(2 votes)

- So when do we use standard change in free energy? What does standard state mean?(1 vote)
- What if the temperature goes up and the remaining factors remain the same, there's no effect on delta G? or maybe if temperature goes up, delta G naught can't be used in the equation? or the equation is not true if temperature changes? And the equation for delta G is only for standard room temperature?(1 vote)
- How did you get -15.9 for the last answer i keep getting -25.7(1 vote)
- at around the1:15mark, jay says that as long as there's a difference in free energy between the reactants and the products the net reaction will proceed to the left or right.'

I'm confused by why the difference between the free energy between the reactants and products would change as the reaction moves towards equilibrium as it's the same reaction?

In general,I'm confused by why ΔG changes as the reaction goes on. I i know the equation ΔG == ΔG° + RTlnQ exists, but i would prefer a more intuitive/qualitative reason if possible.

i don't understand why the change in enthalpy or entropy could change for a a specific reaction, so my current guess is that temperature changes...?(1 vote)- Having a nonzero value for the change in Gibbs free energy mean either the products or the reactants have more free energy than the other. If ΔG is negative, the products have less free energy than the reactants and so the reaction speeds up the forward reaction to create more products (since reactions want to minimize their free energy). If ΔG is positive, then the reactants have less free energy than the products and the reverse reaction speeds up to create more reactants. One when ΔG is zero do neither the forward or reverse reactions speed up, and instead they remain constant. This only happens when the chemical concentrations are at equilibrium, and the reverse and forward reactions being equal is the kinetic way of defining equilibrium.

ΔG changes during a reaction because the reaction quotient, Q, changes. The reaction quotient is essentially a measure of how far are the chemical’s concentrations from the ideal equilibrium concentrations. And since reactions want to attain equilibrium, ΔG is also a measure of how strongly a reaction wants to reach equilibrium. So as the reaction progresses, Q approaches K and therefore ΔG also approaches 0.

Gibbs free energy is a deeper thermodynamic way to explain why equilibrium behaves the way it does.

Hope that helps.(1 vote)

## Video transcript

- [Instructor] Understanding the concept of nonstandard free energy changes is really important when it
comes to a chemical reaction. For this generic chemical reaction, the reactants turn into the products. And nonstandard free energy
change is symbolized by delta G. And notice, this delta G doesn't have the naught superscript, and therefore, it's nonstandard
change in free energy. The reason why this
concept is so important for a chemical reaction is by calculating the nonstandard change in free energy, we can figure out which direction the net reaction will proceed. When delta G is less than zero, so when delta G is negative, the forward reaction is
thermodynamically favored. Therefore, the reactants
will turn into the products. And the amount of products will increase, and the amount of reactants will decrease. When delta G is greater than zero, so when delta G is positive, the forward reaction is not
thermodynamically favored, which means the reverse
reaction is favored. So the net reaction goes
to the left to increase the amount of reactants and
decrease the amount of products. As long as there's a
difference in free energy between the reactants and the products, the net reaction will
proceed either to the left or to the right. However, when there's no
difference in free energy between the reactants and the products, or delta G is equal to zero, the reaction is at equilibrium. And when the reaction is at equilibrium, the concentrations of reactants and products remain constant. So it's useful to think
about nonstandard change in free energy as the driving force for a chemical reaction. As long as there's a
difference in free energy between reactants and products, the net reaction will move
one direction or the other to the left or to the right. However, when there's no
difference in free energy between reactants and products, there's no more driving force and the reaction is at equilibrium. Next, let's look at the equation that relates nonstandard
change in free energy, delta G, to standard change in free
energy, delta G naught. Remember, the superscript naught refers to the substances being
in their standard states under a pressure of one atmosphere. So delta G, or the nonstandard
change in free energy, refers to the instantaneous difference in free energy between the
reactants and the products. So when that reaction moves
to the left or to the right, this value is always changing. It's the driving force for the reaction. Delta G naught is a different situation. Delta G naught is a constant
at a certain temperature. And that's because it's
referring to the difference in free energy between
reactants and products in their standard states. And so this value will remain the same as the net reaction moves
to the left or to the right. To calculate the nonstandard change in free energy, delta G, it's equal to the standard change in free energy, delta G naught,
plus RT natural log of Q, where R is the ideal gas constant, T is the temperature in Kelvin, and Q is the reaction quotient. Next, let's calculate delta
G for a chemical reaction. And this is the reaction for
the synthesis of ammonia gas from nitrogen gas and hydrogen gas. Our goal is to calculate
delta G at 25 degrees Celsius at the moment in time
when the partial pressures of all three gases are one atmosphere. So one atmosphere for
nitrogen, for hydrogen, and for ammonia. At 25 degrees Celsius, delta
G naught for this reaction is equal to negative 33.0
kilojoules per mole of reaction. So our goal is to calculate delta G. And we know delta G naught, but we need to figure out what
Q is at this moment in time, when the partial
pressures of all the gases are one atmosphere. We can get the expression
for the reaction quotient Q from the balanced equation. So it would be the partial
pressure of ammonia. And since there's a two,
it'd be the partial pressure of ammonia raised to the second power divided by the partial
pressure of nitrogen. And since there's a one as a coefficient in the balanced equation, it's the partial pressure
of nitrogen raised to the first power, times the partial pressure of hydrogen. And since we have a three here, it'd be the partial
pressure of hydrogen raised to the third power. Since all three gases
have a partial pressure of one atmosphere, when
we plug in one atmosphere, we find that Q is equal to one. And the natural log of
one is equal to zero. So this second term here in
our equation goes to zero. And at this moment in time, delta G is equal to delta G naught. And since delta G naught is equal to negative 33.0 kilojoules
per mole of reaction, since delta G is negative
at this moment in time, the reaction is thermodynamically favored in the forward direction. So the reactants will
react together to form more of the products. The reason why delta G is
equal to delta G naught at this moment in time is because
our reactants and products are all in their standard states. By convention for a gas, standard state refers to
the pure gas at a pressure of one atmosphere. And since all of the
partial pressures are equal to one atmosphere, that
gave us Q is equal to one, which made the second term equal to zero. So only when the substances
are in their standard states is delta G equal to delta G naught. So if we had a different
set of partial pressures, delta G wouldn't be
equal to delta G naught. Let's calculate delta
G for the same reaction at 25 degrees Celsius. However, this time, instead
of having partial pressures of one atmosphere, all three
gasses at this moment in time have a partial pressure
of two atmospheres. Let's start by calculating
the reaction quotient Q. So when we plug in two atmospheres for the partial pressures of our gasses, we find that Q is equal to .25. Next, to calculate delta G,
we need to plug everything into our equation. So delta G naught is still equal
to negative 33.0 kilojoules per mole of reaction. So we can see, we plugged that in here. The ideal gas constant R is 8.314 joules per Kelvin mole of reaction. But to keep our units
the same in kilojoules, you can see I've converted
the ideal gas constant into kilojoules per
Kelvin mole of reaction. So it's .008314. We said the temperature
was 25 degrees Celsius. So 25 plus 273 is equal to 298 Kelvin. And the reaction quotient
Q is equal to .25. So be the it'd natural log of .25. So Kelvin will cancel out. And when we do the calculation, we find that delta G is equal to negative 36.4 kilojoules
per mole of reaction. Since delta G is negative
at this moment in time, the reaction is
thermodynamically favorable in the forward direction. So the net reaction will move to the right to make more products, and to decrease the amount of reactants. An increase in the amount
of ammonia means an increase in the partial pressure of ammonia. And a decrease in the amount of reactants means a decrease in the partial pressures of the reactants. Therefore, when the net
reaction goes to the right, the reaction quotient Q increases. So we started with a relatively
small value for Q of .25. And we know that Q is going to increase. Let's just jump ahead to a point, and let's say that Q is equal
to 1.0 times 10 to the third, and let's plug that into our equation and see what happens to delta G. So delta G naught stays the same at negative 33.0 kilojoules
per mole of reaction. The ideal gas constant's the same. The temperature's the same. So all we've done is increase Q. Once again, Kelvin cancels out, and we find that delta G, the instantaneous
difference in free energy, is equal to negative 15.9
kilojoules per mole of reaction at this moment in time. Since delta G is negative, once again, the forward reaction is favored. So the reaction moves to the
right to increase products and to decrease reactants. So again, Q will continue to increase. And think about what's
happening to delta G. It was negative 36.4, and
now it's negative 15.9. And as Q keeps increasing,
delta G keeps getting closer and closer to zero. Q will keep increasing until
delta G is equal to zero and the reaction is at equilibrium. So as long as Q is not equal
to the equilibrium constant K for this reaction, the
reaction is not at equilibrium. But eventually, Q will be equal to the equilibrium constant K. And if we were to plug that
value into our equation, we would find that delta
G is equal to zero. And therefore, the
reaction is at equilibrium. Let's summarize the difference between the nonstandard change
in free energy, delta G, and the standard change in
free energy, delta G naught. Delta G naught is talking
about the difference in free energy between
reactants and products when reactants and products
are in their standard states. And since that value is a constant, if the temperature is constant, notice for all of our calculations, delta G naught was equal to negative 33.0 kilojoules
per mole of reaction. Delta G is talking about
the instantaneous difference in free energy between
reactants and products. And as long as there's a
difference in free energy, there's a driving force
for the net reaction to go to the left or to the right. And eventually, when
delta G is equal to zero, the reaction is at equilibrium. Delta G is only equal to delta G naught when the reactants and products are in their standard states.