Energy from ATP hydrolysis drives many nonspontaneous cell...

Energy from ATP hydrolysis drives many nonspontaneous cell reactions: $$ \begin{array}{r} \mathrm{ATP}^{4-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{ADP}^{3-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \\ \Delta G^{\circ \prime}=-30.5 \mathrm{~kJ} \end{array} $$ Energy for the reverse process comes ultimately from glucose metabolism: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Find $K$ for the hydrolysis of ATP at $37^{\circ} \mathrm{C}$. (b) Find $\Delta G_{\text {rxn }}^{\circ \prime}$ for metabolism of 1 mol of glucose. (c) How many moles of ATP can be produced by metabolism of $1 \mathrm{~mol}$ of glucose? (d) If $36 \mathrm{~mol}$ of $\mathrm{ATP}$ is formed, what is the actual yield?

Energy from ATP hydrolysis drives many nonspontaneous cell reactions:
$$
\begin{array}{r}
\mathrm{ATP}^{4-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{ADP}^{3-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \\
\Delta G^{\circ \prime}=-30.5 \mathrm{~kJ}
\end{array}
$$
Energy for the reverse process comes ultimately from glucose metabolism:
$$
\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)
$$
(a) Find $K$ for the hydrolysis of ATP at $37^{\circ} \mathrm{C}$.
(b) Find $\Delta G_{\text {rxn }}^{\circ \prime}$ for metabolism of 1 mol of glucose.
(c) How many moles of ATP can be produced by metabolism of $1 \mathrm{~mol}$ of glucose?
(d) If $36 \mathrm{~mol}$ of $\mathrm{ATP}$ is formed, what is the actual yield?

Key Concepts

Stoichiometry and Energy Yield Calculations

Stoichiometry in biochemical reactions involves tracking the molar relationships between reactants and products to determine energy yield and efficiency. This principle is used to calculate how the energy released from one process (such as glucose metabolism) can be partitioned to drive the synthesis of energy carriers like ATP, as well as determine the yield percentage of the produced ATP.

Cellular Respiration and Metabolic Energetics

This key concept encompasses the energy-releasing process of oxidizing substrates (e.g., glucose) to produce useful chemical energy, typically in the form of ATP. It represents the overall conversion of food energy into the usable energy currency within cells, enabling various cellular functions.

Reaction Coupling in Biochemistry

Reaction coupling involves linking an energetically unfavorable (endergonic) reaction to a favorable (exergonic) one, such as ATP hydrolysis, so that the overall process becomes spontaneous. This mechanism is critical in cellular metabolism, allowing nonspontaneous reactions to proceed by harnessing energy from reactions like glucose oxidation.

Gibbs Free Energy and Equilibrium Constant

This concept explains the quantitative relationship between the free energy change (?G°') of a reaction and its equilibrium constant (K) using the equation ?G°' = -RT ln K. It is fundamental in thermodynamics, as it helps predict the position of equilibrium for a reaction and determines whether a reaction is spontaneous under standard conditions.

Transcript

00:01 All right.

00:02 So for part a, we are given the equation for the hydrolysis of atp, and we are given that its delta g is 30 .5 kilojoules.

00:12 So as always, we're going to use our equation here, delta g equals rt, lnk, and we're just going to plug in values.

00:20 Again, always know that r is 8 .34, or 8 .314, excuse me, joules per, calvin moore.

00:39 And because this is in jewels, we need to change our delta g into jewels.

00:45 So, whoops, by multiplying by 1 ,000 to get jewels, you would get 30 ,500 joules.

01:01 So now let's just plug that into our equation.

01:04 So 30 ,500 joules equals negative 8 .314 times 200.

01:25 Whoops.

01:29 Times 298 .98.

01:35 Kelvin m times l .n.

01:41 10 k.

01:44 So if you multiply these together and then divide this by that value, and then you can take e to whatever number you get, you will find k.

01:58 So if you want to take a minute and do those calculations, you should get a k of 1 .38 times 10 to the fifth.

02:11 Perfect.

02:12 All right.

02:13 So the next part asks us to find delta g of reaction for the metabolism of glucose.

02:24 So we're given the reaction and we're going to use this formula, products minus reactants.

02:32 I think it's always helpful to write out kind of a little skeleton and then you can just plug in values where you need to go.

02:38 So the products we have, we're going to need the delta g of co2 plus the delta g of h2 minus the delta g of o2 plus the delta g of glucose.

02:50 Remember, whenever there are coefficients in the reaction, you need to put them in this equation as well.

02:58 So you're going to have six times the delta g of co2 plus six times the delta g of h2 and six times the delta g of h2.

03:11 It's very important to remember that.

03:14 So you can find the delta g of reactions, most likely in your textbook.

03:17 You can also google them, but i have provided them right here.

03:21 So from here, it's just a plug and chug.

03:23 So you're going to put this value here...

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