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## AP®︎/College Chemistry

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

Lesson 6: Coupled reactions# Coupled reactions

AP.Chem:

ENE‑5 (EU)

, ENE‑5.B (LO)

, ENE‑5.B.2 (EK)

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 Δ

*G*° 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)

## Video transcript

- [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.