A thermodynamically unfavored reaction can be driven by coupling it to a favored reaction through one or more shared intermediates. The sum of the two reactions yields an overall reaction that has a negative ΔG° value. Created by Jay.
Want to join the conversation?
- Is it possible to take H2SO4 and decompose it into H2O and SO3?(1 vote)
- Yes, it's possible to decompose sulfuric acid into water and sulfur trioxide at high temperatures around 300 - 450 °C.
Hope that helps.(3 votes)
- How would you go about adding together "Delta G naught is greater than 0" + "Delta G naught is much less than 0" to get the sum "Delta G naught is less than 0." Intuitively, how does this work?
--Forever Learner--(1 vote)
- Using the inequalities like that is symbolically saying a value is positive or negative in math. If it is greater than zero then it is positive, if it less than zero then it is negative. Using more than one inequality sign signifies it is much greater or much lesser.
It's essentially saying that if you add a reaction with a slightly positive free energy to a reaction with a very negative free energy the resulting sum will be a negative free energy value.
Hope that helps.(1 vote)
- [Instructor] Coupled reactions use a thermodynamically favorable reaction to drive a thermodynamically unfavorable reaction. For example, let's look at a hypothetical reaction where reactants A and B combine to form products C and i. The standard change in free energy for this reaction, delta G naught, let's say, is greater than zero. And when delta G naught is greater than zero, that's a thermodynamically unfavorable reaction. And at equilibrium, the equilibrium constant, K, is less than one. When K is less than one, that means there are more reactants than products at equilibrium. And if our goal is to make a lot of the product C, that won't happen as long as this reaction is thermodynamically unfavorable. Next, let's look at another hypothetical reaction where i reacts with D to form E and F as the products. For this hypothetical reaction, the standard change in free energy, delta G naught, is much less than zero. Since delta G naught for the second hypothetical reaction is less than zero, reaction two is thermodynamically favorable. And at equilibrium, the equilibrium constant, K, is much greater than one. If K is much greater than one, then we can assume that the second reaction essentially goes to completion. So all the i and the D will react together to form E and F. And at equilibrium, there are way more products than there are reactants. Remember that our goal was to make a lot of the product C, but we can't get that by just running the first reaction by itself. However, by coupling the first reaction, which is thermodynamically unfavorable, to the second reaction, which is thermodynamically favorable, we can make a significant amount of product C. These two hypothetical reactions can be coupled together because they share a common intermediate, i. Even though reaction one is thermodynamically unfavorable, even if it forms only a small amount of i, since reaction two is thermodynamically favorable, the i from reaction one will be used up in reaction two to form the products for reaction two. Removing the product i causes the equilibrium to shift to the right in the first reaction, therefore producing more of our desired product C. So the thermodynamically favorable reaction is driving the thermodynamically unfavorable reaction to the right. Since we are coupling these two reactions together, if we add reaction one to reaction two, the intermediate would cancel out. And for the reactants, we would get A plus B plus D; and for the products, we would get C plus E plus F. To find delta G naught for this overall equation, we would need to add together delta G naught for reaction one and delta G naught for reaction two. So delta G naught for this overall equation would be less than zero, which tells us this overall equation is thermodynamically favorable. And the equilibrium constant is greater than one, which means at equilibrium, we'll have a decent amount of our desired product C. Next, let's look at a practical example of coupled reactions. Let's say that our goal is to extract solid copper from copper(I) sulfide. The other product of this reaction is sulfur. However, delta G naught for this reaction is equal to positive 86.1 kilojoules per mole of reaction. Since delta G naught is positive, this reaction is thermodynamically unfavorable. And at equilibrium, there'd be very little of our product, so we wouldn't get very much copper. The solution is to couple this first reaction to a thermodynamically favorable reaction. And that's the conversion of sulfur into sulfur dioxide. So solid sulfur plus oxygen gas give sulfur dioxide, and delta G naught for this reaction is equal to negative 300.4 kilojoules per mole of reaction. So adding our two reactions together, the common intermediate, sulfur, cancels out, and that gives us copper(I) sulfide plus oxygen for the reactants goes to solid copper and sulfur dioxide for the products. And if we add the two delta G naught values together, positive 86.1 plus negative 300.4 gives a delta G naught for this overall equation equal to negative 214.3 kilojoules per mole of reaction. Since delta G naught is negative, now, this overall reaction is thermodynamically favorable with an equilibrium constant greater than one. Therefore, we will now produce a lot of copper. So burning copper(I) sulfide in air gives us a feasible way of extracting copper from copper(I) sulfide. Coupled reactions are also very important in biochemistry. For example, two amino acids are alanine and glycine. And if these two amino acids come together, they form what's called a dipeptide. It's very important for the body to be able to make dipeptides because if we were to add on some more amino acids, we would form what's called a polypeptide, which eventually leads to the formation of a protein. However, the formation of dipeptides is thermodynamically unfavorable. Therefore, the body needs a way of making this reaction thermodynamically favorable so the body can make proteins. To do this, the body uses ATP, which stands for adenosine triphosphate. And here's the dot structure for ATP in the ionized form. ATP is often called a high energy compound, and the hydrolysis of ATP is used to drive protein synthesis. Next, let's look at the equation showing the hydrolysis of ATP to ADP, which is adenosine diphosphate and inorganic phosphate. The standard change in free energy for the hydrolysis of ATP is equal to negative 31 kilojoules per mole of reaction. Since delta G naught is negative, the hydrolysis of ATP is thermodynamically favorable. Alanine and glycine combine to form the dipeptide alanylglycine and water. Delta G naught for the formation of this dipeptide is equal to positive 29 kilojoules per mole of reaction. Since delta G naught is positive, this reaction is thermodynamically unfavorable. With the help of an enzyme, the unfavorable reaction is coupled to the favorable reaction. And if we add the two reactions together, water would cancel out and give us alanine plus glycine plus ATP for the reactants, and alanylglycine and ADP and inorganic phosphate for the products. Adding the two values of delta G naught together, we get delta G naught is equal to negative two kilojoules per mole of reaction. And since delta G naught is negative, it's now favorable for the body to synthesize this dipeptide.