Thermodynamic and Kinetic Control; Coupled Reactions
A diamond ring sits on your finger for decades. Graphite is the more stable form of carbon, so why doesn't the diamond turn into graphite? And how do your cells build large molecules that, on their own, would never form? Both questions come down to separating two ideas: can a reaction happen, and how fast does it happen.
Favorable does not mean fast
Two different questions describe any reaction.
Thermodynamics asks: is the reaction favorable? The answer comes from $\Delta G^\circ$. If $\Delta G^\circ < 0$, then $K > 1$ and the products are favored at equilibrium.
Kinetics asks: how fast does it go? The answer comes from the activation energy $E_a$ (and the rate law). A large $E_a$ means very few collisions have enough energy, so the rate is extremely slow.
The two are linked to different parts of the energy profile. $\Delta G^\circ$ compares the start and the end. $E_a$ is the height of the hill in between.
In the figure, the label Eₐ stands for the activation energy $E_a$ (the height of the hill), and $\Delta G^\circ$ is the drop from reactants to products.
So a reaction can be favorable but extremely slow. Diamond to graphite has $\Delta G^\circ < 0$ at 298 K and 1 atm, but its $E_a$ is enormous because carbon–carbon bonds must break throughout a rigid network. The diamond is thermodynamically unstable but kinetically stable. Chemists call such a substance metastable.
A catalyst lowers $E_a$ by giving the reaction a different mechanism. It speeds up the forward and reverse reactions equally, so $\Delta G^\circ$ and $K$ do not change. It only gets you to equilibrium sooner.
At 298 K and 1 atm, the conversion of diamond to graphite has $\Delta G^\circ < 0$, yet a diamond ring does not turn into graphite on your finger. Is the conversion…
At 298 K and 1 atm, the conversion of diamond to graphite has $\Delta G^\circ < 0$, yet a diamond ring does not turn into graphite on your finger. Is the conversion thermodynamically favorable? Is it kinetically accessible at room temperature? Name the quantity that explains why no change is observed.
Label each statement as describing thermodynamic favorability (T), kinetic accessibility (K), or neither (N): (i) The reaction has a very large activation energy. (ii)…
Label each statement as describing thermodynamic favorability (T), kinetic accessibility (K), or neither (N):
(i) The reaction has a very large activation energy.
(ii) The reaction has $\Delta G^\circ = -250\ \text{kJ/mol}$.
(iii) A catalyst makes the reaction reach equilibrium within seconds.
(iv) The reaction has $K \gg 1$.
A mixture of $\text{H}2(g)$ and $\text{O}2(g)$ at room temperature can sit for years without visible change, even though…
A mixture of $\text{H}_2(g)$ and $\text{O}_2(g)$ at room temperature can sit for years without visible change, even though $2\,\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\,\text{H}_2\text{O}(l)$ has $\Delta G^\circ \approx -474\ \text{kJ}$. A spark, or a bit of platinum powder, makes it react at once.
(a) Explain the lack of reaction in terms of thermodynamics and kinetics. (b) State how the platinum catalyst changes $E_a$, $\Delta G^\circ$, and $K$. (c) Explain what the spark does.
Thermodynamic and kinetic products
Sometimes a reactant R can go to two different products, A or B, by competing pathways. Which one you get depends on the conditions.
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The kinetic product forms fastest. Its pathway has the lower $E_a$.
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The thermodynamic product is the most stable. It has the more negative $\Delta G^\circ$ and dominates if the system reaches equilibrium.
Often they are different substances: the faster route leads to a less stable product.
Temperature decides which product you collect:
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Low temperature: few molecules can climb the high barrier, and the reaction to A is mostly irreversible. A forms faster and stays. The reaction is under kinetic control.
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High temperature: molecules have enough energy to cross both barriers, and A can even go back to R and over to B. Given time, the system settles into the most stable mixture. The reaction is under thermodynamic control.
A reactant R can form product A or product B. The pathway to A has $E_a = 50\ \text{kJ/mol}$ and $\Delta G^\circ = -20\ \text{kJ/mol}$. The pathway to B has…
A reactant R can form product A or product B. The pathway to A has $E_a = 50\ \text{kJ/mol}$ and $\Delta G^\circ = -20\ \text{kJ/mol}$. The pathway to B has $E_a = 90\ \text{kJ/mol}$ and $\Delta G^\circ = -45\ \text{kJ/mol}$.
(a) Which product forms faster? (b) Which product is more stable and would dominate if the system reached equilibrium? (c) Identify the kinetic and the thermodynamic product. (d) Explain why a low-temperature experiment and a high-temperature experiment might give different product mixtures.
For $\text{C(graphite)} \rightarrow \text{C(diamond)}$ at 298 K, $\Delta H^\circ = +1.9\ \text{kJ/mol}$ and $\Delta S^\circ = -3.4\ \text{J/(mol·K)}$. (a) Calculate…
For $\text{C(graphite)} \rightarrow \text{C(diamond)}$ at 298 K, $\Delta H^\circ = +1.9\ \text{kJ/mol}$ and $\Delta S^\circ = -3.4\ \text{J/(mol·K)}$.
(a) Calculate $\Delta G^\circ$ at 298 K. (b) Using the signs of $\Delta H^\circ$ and $\Delta S^\circ$, explain whether there is any temperature at which this conversion becomes favorable at standard conditions. (c) Synthetic diamonds are made at very high pressure; explain why pressure helps, given that diamond is denser than graphite. (d) Explain why natural diamonds do not convert to graphite even though the reverse process is favorable.
Using temperature and pressure: the sign of $\Delta G^\circ$
Recall $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Always convert so units match: $\Delta S^\circ$ is in J/(mol·K), while $\Delta H^\circ$ and $\Delta G^\circ$ are in kJ/mol.
For graphite → diamond at 298 K, $\Delta H^\circ = +1.9\ \text{kJ/mol}$ and $\Delta S^\circ = -3.4\ \text{J/(mol·K)}$:
$$\Delta G^\circ = 1.9\ \text{kJ/mol} - (298\ \text{K})(-0.0034\ \text{kJ/(mol·K)}) = 1.9 + 1.0 = +2.9\ \text{kJ/mol}$$$\Delta H^\circ > 0$ and $\Delta S^\circ < 0$ make $\Delta G^\circ$ positive at every temperature. No temperature makes the conversion favorable at standard conditions. Diamond is made by changing the conditions instead: at very high pressure, the denser form is favored, because diamond takes up less volume than graphite. Note the roles: the high pressure is what changes which form is favorable. The high temperature and a molten metal catalyst only provide kinetic access, so the carbon rearranges in a reasonable time. (Standard conditions, 1 atm, are where $\Delta G^\circ$ is positive.)
Coupled reactions
An unfavorable reaction ($\Delta G^\circ > 0$) can still happen if it is coupled to a more favorable one that shares a species. Add the two reactions, cancel what appears on both sides, and add their $\Delta G^\circ$ values.
Cells do this constantly with ATP hydrolysis, $\Delta G^\circ = -30.5\ \text{kJ/mol}$. In real cells the coupling usually happens through a shared intermediate: one step transfers a phosphate from ATP onto a reactant, and a second step releases it.
Notice that coupling does not break any rule: the overall $\Delta G$ is negative. The unfavorable step is not "forced"; it is part of an overall favorable process.
Coupling changes $K$ dramatically. For an uncoupled reaction with $\Delta G^\circ = +25.0\ \text{kJ/mol}$ at 298 K, $K = e^{-25000/(8.314 \times 298)} \approx 4\times10^{-5}$. Coupled to ATP hydrolysis, $\Delta G^\circ = -5.5\ \text{kJ/mol}$ and $K \approx 9$. That is an increase of about a factor of $2\times10^{5}$.
In cells, the phosphorylation of glucose is coupled to ATP hydrolysis: $\text{glucose} + \text{P}i \rightarrow \text{glucose-6-phosphate} + \text{H}2\text{O}$,…
In cells, the phosphorylation of glucose is coupled to ATP hydrolysis:
$\text{glucose} + \text{P}_i \rightarrow \text{glucose-6-phosphate} + \text{H}_2\text{O}$, $\Delta G^\circ = +13.8\ \text{kJ/mol}$
$\text{ATP} + \text{H}_2\text{O} \rightarrow \text{ADP} + \text{P}_i$, $\Delta G^\circ = -30.5\ \text{kJ/mol}$
Calculate $\Delta G^\circ$ for the coupled process and state whether it is thermodynamically favorable.
Copper metal can be obtained from copper(I) sulfide by coupling two reactions at 298 K: $\text{Cu}2\text{S}(s) \rightarrow 2\,\text{Cu}(s) + \text{S}(s)$,…
Copper metal can be obtained from copper(I) sulfide by coupling two reactions at 298 K:
$\text{Cu}_2\text{S}(s) \rightarrow 2\,\text{Cu}(s) + \text{S}(s)$, $\Delta G^\circ = +86.2\ \text{kJ}$
$\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g)$, $\Delta G^\circ = -300.1\ \text{kJ}$
(a) Write the net reaction. (b) Calculate the net $\Delta G^\circ$. (c) Explain how the species shared by both reactions makes the first process possible, and what a very negative $\Delta G^\circ$ implies about $K$.
A biochemical conversion $\text{X} + \text{P}i \rightarrow \text{Y} + \text{H}2\text{O}$ has $\Delta G^\circ = +25.0\ \text{kJ/mol}$ at 298 K. A cell couples it to ATP…
A biochemical conversion $\text{X} + \text{P}_i \rightarrow \text{Y} + \text{H}_2\text{O}$ has $\Delta G^\circ = +25.0\ \text{kJ/mol}$ at 298 K. A cell couples it to ATP hydrolysis, $\Delta G^\circ = -30.5\ \text{kJ/mol}$, through a shared intermediate: $\text{X} + \text{ATP} \rightarrow \text{X-P} + \text{ADP}$, followed by $\text{X-P} + \text{H}_2\text{O} \rightarrow \text{Y} + \text{P}_i$.
(a) Calculate $\Delta G^\circ$ and $K$ for the coupled process, and $K$ for the uncoupled reaction. Use $R = 8.314\ \text{J/(mol·K)}$. (b) By roughly what factor does coupling increase $K$? (c) Aaliyah says, "ATP just adds energy to the reaction." Evaluate this idea and explain why a shared intermediate is needed. (d) Explain what would happen to the energy of ATP hydrolysis if it occurred with no coupling.
Summary
Back to the start: the diamond is stuck by a huge $E_a$, not by thermodynamics, and cells get unfavorable reactions to run by coupling them to ATP.