<p>I still don’t get the 2008 #8(d)(ii) ahah… You scholars said that the reduction with H2O was more likely to occur because it had a higher electrical potential than Na. But O2 has an even higher reduction potential e-, as stated in the first row of the table. I don’t get it at all…</p>
<p>But here’s how I originally came to the wrong conclusion that Na was being reduced:
I knew that the electric potential (E) had to be negative b/c current was being provided to the cell (thus it was an electrolytic cell). So I just shifted the Na/I2 equations around so it made the E negative, which was I2 being the oxidation reaction and Na the reduction reaction. So basically the Na/I2 combo and the H2O/I2 combo both make the reaction negative, but what I don’t get how you distinguish which one to employ.</p>
<p>Abrayo: The Q (where the reaction is at that moment) is less than Kp. That means it has to proceed further towards the products in order to reach Kp. I think you’re a little bit confused as to what Q is standing for, as you seem to be thinking that it is another K value.
One way I think of whether the reaction will continue forwards or backwards when looking at Q vs K is the “number line” trick.
<a href=“http://img839.imageshack.us/img839/2937/qvsk.gif[/url]”>http://img839.imageshack.us/img839/2937/qvsk.gif</a></p>
Since on the left we have a solid and ions, and on the right we have a liquid, a gas, and ions, entropy must have increased. I like to think of it as energy is being put into the molecules making them more agitated/jittery/random kinda like giving a 5 year old red bull and entropy is a measure of randomness.
First before anything, you consider states of matter is what my chem teacher told me.</p>
<p>Orangefoot: delta S is positive because there are2 moles of gas in the product side and none on the reactants, gases are the most spontaneous of the states so it is a positive S.</p>
<p>Ahh, so for the question I posted:
Since Q < K that means that once it proceeds, more products will be formed . . . and since the fraction must be constant, an increase in numerator means that the denominator must decrease, right?</p>
<p>On a plus side, I got a 5 on the 2008 exam!</p>
<ol>
<li>How do you tell if a molecule is polar or non polar?!!!</li>
<li>Why does the boiling point of the Nitrogen family increase as we go down the periods?</li>
</ol>
<p>orangefoot: there are more moles of gas in the products than in the reactants, that’s why. As for halving the rate of the reaction, it’s asking for the rate of the whole reaction, which has a coefficient of one… you just have to get rid of the two moles of NO2 because it’s appearing twice as fast as something with a coefficient of one. Sorry if that was a bad explanation.</p>
<p>Draw the LDS. Look for any lone pairs. If you see some, it’s polar. Also, look at the atoms that make the bonds up. N, O, and F are very electronegative and pull. However, if the other bonds are also made of those same elements and there are no lone pairs(SF4 polar vs SF6 non-polar), then its non-polar.</p>
<ol>
<li>How do you tell if a molecule is polar or non polar?!!!</li>
<li>Why does the boiling point of the Nitrogen family increase as we go down the periods?</li>
<li>Why is hexane less soluble than glucose?</li>
</ol>
<ol>
<li>Most likely London Dispersion Forces that increase as atomic size increases (which does as n gets larger where n=period)</li>
<li>Draw the LDS. Glucose has some O-H groups on it which means dipole-dipole interactions and hydrogen bonding which when combined = higher solubility in polar H2O versus non-polar benzene which relies solely on much weaker LDF</li>
</ol>
<p>Why isn’t it C? It says it’s A.
If deltaH> 0 then it’s endothermic, right? So heat is a reactant.
Therefore adding heat would shift it to the right, and produce more HI?</p>
<p>And 36… shouldn’t it be 6.02 times 10^22, because that’s o.1 moles?</p>
You only have H2O and the Na+</p>![http://img839.imageshack.us/img839/2937/qvsk.gif[/url]](http://img839.imageshack.us/img839/2937/qvsk.gif%5B/url%5D){kind=link}
