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Calculate $\Delta G$ for the reaction $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$ at $25^{\circ} \mathrm{C}$ for the following conditions:(a) $\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} \mathrm{M},\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-7} \mathrm{M}$(b) $\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M},\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-4} \mathrm{M}$(c) $\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-12} \mathrm{M},\left[\mathrm{OH}^{-}\right]=2.0 \times 10^{-8} \mathrm{M}$(d) $\left[\mathrm{H}^{+}\right]=3.5 \mathrm{M},\left[\mathrm{OH}^{-}\right]=4.8 \times 10^{-4} \mathrm{M}$

scc: $1 ;$ bcc: $2 ;$ fcc: 4.

Chemistry 102

Chapter 17

Entropy, Free Energy, and Equilibrium

Thermodynamics

Rice University

University of Maryland - University College

University of Kentucky

University of Toronto

Lectures

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So we have these conditions for the auto ionization of water, different concentrations of ID, rhodium and hydroxide. And we want to calculate the Delta G for these various conditions. And ah, so we need to know the value of Delta G zero for the reaction, first of all. And so I plugged in the value of the K W one times 10 to the minus 14 into the equation for calculating Delta G zero from K and the value of our and I'm assuming we're here at 25 degrees. Yes, we are. That's where this value occurs. And so, uh, when I when I do the calculation, I get about 80 killer jewels Permal positive for this reaction. So Ah, that means it doesn't go very far to the right, as we will know. And so that's our Delta G zero and, ah, we want to calculate Delta G for these various conditions. So the equation for that is that the Delta G is equal to the Delta G zero plus r t l and Q's. So we I've gone ahead and calculated queue for each of these different situations. Uh, when both the hydro knee my on and the hydroxide er attended the minus seven. I get one times 10 to the minus 14. Well, I don't need to calculate to know that the value is zero in that case, because the reaction is at equilibrium already. Ah, that is thieve Al you of Q is equal to the value of K. That is one of the definitions of equilibrium. So I know the value of Delta G under that condition. Now, in the second set, Q is one times 10 to the minus seven. Selimi, go ahead and start plugging in and calculating. And I've saved this number for rt since I'm gonna have to calculated over and over again, um, so that I only have to multiply by the l n Q. And so let me do that. So in this case, are Q is a negative 16. Basically, uh, I'm sorry. The log of Q Q is one times 10 to the minus seven. The natural long of Q is one times is negative, 16 roughly and then multiplying that, uh, I get from the 10 to the minus seven. I get negative. 39.9. I killed Joe sperm role. And so that means, as we know, that is a, uh Do I know? I'm sorry That that is the value. I apologize. That is the value. Uh, RTL and Q and I have to add to that Delta G zero and then I get a positive 39.96. Really? That's 40 eso highly unfavourable to G O in the four direction as we know que here is quite large. And so Q has to decrease. And that means the reaction has to go to the left. Eso zero for part a 44 part B, part C. Now we've got a very small value, and so again, that is gonna be okay, said log of Q This time his, uh negative 45 and multiplying that by r t, I get, uh, negative 112 0.4 killer jewels for minus RTL on cue. And so I get a negative 32 0.5 killer jewels. So with this very low value of Q, I've got very little of Thea of the age plus in the hydroxide. Ah, relative to what I would have it equilibrium. And so the reaction is favorable to go forward and produce more. And and so that is with 10 to the minus, two times 10 to the minus 20. And then finally, let me write about this time. So first of all, that log is just a negative six. And so that's gonna be around 15 killer jewels when I multiply. And that would be negative. 15.6. You know, Jules, for the rt lo Ellen Q and then adding the 79 49 gives me a big positive 64 0.3 killer jewels promote. So here we've got very high age plus in hydroxide related to what we would have at equilibrium. And so the reaction will go quite a bit to the left. I had a very large positive, uh, Delta G. And so it goes back to water the H plus and the hydroxide combining to form water.

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