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Current time:0:00Total duration:10:15
AP.Chem:
ENE‑5 (EU)
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ENE‑5.A.2 (EK)

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.
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