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Problem 104

A key step in the metabolism of glucose for energy is the isomerization of glucose-- 6 -phosphate $(\mathrm{G} 6 \mathrm{P})$ to fructose 6 -phosphate $(\mathrm{F} 6 \mathrm{P}) : \mathrm{G6P} \quad \rightleftharpoons \mathrm{F6P} ; K=0.510$ at 298 $\mathrm{K}$

(a) Calculate $\Delta G^{\circ}$ at 298 $\mathrm{K}$ .

(b) Calculate $\Delta G$ when $Q$ , the $[\mathrm{F} 6 \mathrm{P}] /[\mathrm{G} 6 \mathrm{P}]$ ratio, equals $10.0 .$

(c) Calculate $\Delta G$ when $Q=0.100$ .

(d) Calculate $Q$ if $\Delta G=-2.50 \mathrm{kJ} / \mathrm{mol}$

Answer

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## Discussion

## Video Transcript

we're given the equilibrium constant K for the isomerization of glucose six phosphate into fruit coast six phosphate. And we want to use this information to help us solve out the appropriate values that are required for parts. Eighth your d of this problem and these two equations will be useful for us as we do that and part A. We want to find the changing Gibbs free energy at standard conditions, using the equilibrium constant and a temperature of 298 Kelvin. So therefore, we need to use the equation on the top to help us do that. So that would be equal to negative our which is 8.314 Jules per mole times Kelvin times the temperature of 298 kelvin times the natural log that equilibrium K constant, which is given a 0.510 When we work through the math, we should get a final answer in part a really changing Gibbs free energy at standard conditions. For this, I saw memorization reaction at 298 Kelvin to be 1.67 times 10 to the third. And when we can't sof units of Kelvin. We're left with finally units of jewels per mole. So that is the answer for part A and part B. We we need to find Delta G, and we're told that your issue of lactose six phosphate to glucose six phosphate is 10 to 1, so concentration of fructose six phosphate through the concentration of glucose six phosphate equals 10. We write out the equilibrium, constant expression. You see that we need the concentration of the product, which is F six p by two by the concentration of the reactant which is G six p. And when K is equal to this value and we're at equilibrium. But in per p, it's not equal to that value. It's equal to 10. So that means that that represents que reaction question. In this case, we need to find Delta G it non standard conditions in part B and we're giving Q. So we need to use that equation this time. So Delta G equals Delta G at standard conditions, which is what we just saw. Four in part A. Because this symbol denotes standard conditions. But now we have moved to non standard conditions. Since we have a value that for the equilibrium constant that deviates from the true value. So now we have a reaction quotient. Hugh instead of K. So one point 67 times 10 to the third jewels per mole again was that Delta G value at standard conditions. Now we add our tln of Q or is 8.314 Jules per mole. I'm skilled in times t 298 killed in times the Ellen of Q, which we're told in this be that that is equal to 10. And now when we work through the math, get Delta G equals 7.37 times 10 to the third. That again will have units of jewels per mole actor re cancel out units of Kelvin and now imports. See, we're essentially doing the same thing is Part B were explicitly given a value for Q. It's not written in a ratio of the concentrations of product to the reactant, so we use that value of Q and the same equation, using the Delta G at standard conditions from Part A again 1.67 times 10 to the third joules per mole plus RTL in of K 8.314 joules per mole kelvin Times t, 298 k Times The natural log of Q and import. See, we're told that Q equals 0.100 when you plug in those numbers into this equation and get Delta G and non standard conditions and pert C equals negative four 0.4 times 10 to the third and again it has units of jewels per mole and in Part D. This time we're told what Delta G is a non standard conditions, and we want to know what Q is so we can rearrange that equation this time in order to solve for Q. We do that. We see that Ellen of Q is equal to Delta G at non standard conditions, which is what we're given minus Delta G at standard conditions, which is what we saw it for, in part a divided by R T. So to isolate Q on the left side need to raise each side of the equation to to the e to the power of whatever it is on that side of the equation. And so we do that we see the Q equals E to the power of Delta G at non standard conditions, which is given as negative two point 50 illegals per mole. But we need to convert this into jewels per mole by multiplying by 1000 so that when we subtract Delta G at standard conditions from Part A, which again is 1.67 times 10 to the third, we can subtract that same unit of jewels per mole. And then we divide by R T, which are the same values is we've used before, 8.314 joules per mole times, Kelvin times that temperature of 298 Kelvin. And now, when we saw for Q, we should get the unit lists answer of zero point 19