Test 2 Question 8b
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Test 2 Question 8b
Wednesday in Dane's section
Consider the following unbalanced half reactions
Hg2Cl2(s) + 2e- -> 2Hg(l) + 2Cl-(aq) Enaught = +0.27V
2H+(aq) + 2e- -> H2(g) Enaught = 0.00V
b) If PH2 = 2.0 bar, [H+] = 3.16x10^4 M and [Cl] = 0.75M, what is Q?
For this problem, how to you deal with partial pressure and M? Do you have to convert partial pressure to M before you solve for Q? I remember seeing problems like this in the textbook but the answer key just used both partial pressure and M when solving Q, and I was wondering why since when solving for K or Q in the equilibrium chapter, we had to either convert everything to partial pressure or M
Consider the following unbalanced half reactions
Hg2Cl2(s) + 2e- -> 2Hg(l) + 2Cl-(aq) Enaught = +0.27V
2H+(aq) + 2e- -> H2(g) Enaught = 0.00V
b) If PH2 = 2.0 bar, [H+] = 3.16x10^4 M and [Cl] = 0.75M, what is Q?
For this problem, how to you deal with partial pressure and M? Do you have to convert partial pressure to M before you solve for Q? I remember seeing problems like this in the textbook but the answer key just used both partial pressure and M when solving Q, and I was wondering why since when solving for K or Q in the equilibrium chapter, we had to either convert everything to partial pressure or M
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Re: Test 2 Question 8b
You need to convert bar to M. We need to keep unit the same. Think of the idea gas equation PV=nRT, then we have P/RT=n/V. The number of mole divided by volume is molarity, so P/RT=Molarity. That's how we convert bar to M. Hope it helps!
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Re: Test 2 Question 8b
Qining Jin 1F wrote:Wednesday in Dane's section
Consider the following unbalanced half reactions
Hg2Cl2(s) + 2e- -> 2Hg(l) + 2Cl-(aq) Enaught = +0.27V
2H+(aq) + 2e- -> H2(g) Enaught = 0.00V
b) If PH2 = 2.0 bar, [H+] = 3.16x10^4 M and [Cl] = 0.75M, what is Q?
For this problem, how to you deal with partial pressure and M? Do you have to convert partial pressure to M before you solve for Q? I remember seeing problems like this in the textbook but the answer key just used both partial pressure and M when solving Q, and I was wondering why since when solving for K or Q in the equilibrium chapter, we had to either convert everything to partial pressure or M
Haocheng Zhang 2A wrote:You need to convert bar to M. We need to keep unit the same. Think of the idea gas equation PV=nRT, then we have P/RT=n/V. The number of mole divided by volume is molarity, so P/RT=Molarity. That's how we convert bar to M. Hope it helps!
I asked Dane and he said that you use both partial pressure and concentration when solving for Q, but can someone confirm that?
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Re: Test 2 Question 8b
You can use both partial pressure and concentration in the equilibrium expression as long as the pressure is in bars and the concentration is in molarity.
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Re: Test 2 Question 8b
Haocheng Zhang 2A wrote:You need to convert bar to M. We need to keep unit the same. Think of the idea gas equation PV=nRT, then we have P/RT=n/V. The number of mole divided by volume is molarity, so P/RT=Molarity. That's how we convert bar to M. Hope it helps!
Keliana Hui 2E wrote:You had to assume standard conditions, so temperature = 298 K
Converting it to molarity using this method gives you a different value of Q than the correct answer.
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Re: Test 2 Question 8b
I asked my high school chemistry teacher this question, he said we don't have to convert them to the same unit becasue equilibrium constant is unitless, so I am wrong. Sorry about that.
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Re: Test 2 Question 8b
Haocheng Zhang 2A wrote:I asked my high school chemistry teacher this question, he said we don't have to convert them to the same unit becasue equilibrium constant is unitless, so I am wrong. Sorry about that.
Do we still use molarity and bars then or do we convert it to atm or some other pressure unit?
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