1 consider the reaction ch3oh l ch4 g 12 o2 g a calculate...

1. Consider the reaction: CH3OH (l) -> CH4 (g) + 1/2 O2 (g) (a) Calculate $\Delta H^0_{rxn}$ and $\Delta G^0_{rxn}$ for this reaction at 25°C and 1 atm. (b) Calculate $\Delta S_{rxn}$ for this reaction at 25°C and 1 atm. (c) Calculate $\Delta S_{rxn}$ for this reaction at 37°C and 1 atm. State any assumptions you make. Some information that may be useful is below. These values are given at 25°C and 1 atm. | | $\Delta H^0_f$, kJ/mol | $\bar{S}^0$, J/mol K | $\Delta G^0_f$, kJ/mol | $\bar{C}_p$, J/mol K | |---|---|---|---|---| | CH3OH (l) | -238.57 | 126.8 | -166.23 | 79.5 | | CH4 (g) | -74.81 | 186.264 | -50.72 | 35.69 | | H2O (g) | -241.818 | 188.825 | -228.572 | 33.76 | | H2O (l) | -285.830 | 69.91 | -237.129 | 75.4 | | O2 (g) | 0 | 205.138 | 0 | 29.38 |

Transcript

00:01 So this is our reaction and ch30h becoming carbon monoxide and h2 gas.

00:06 And i already looked up the delta h formation in the states for the entropy for all these reactions and products.

00:15 And so here they are in this table right here.

00:18 And so this is what we're going to use in order to figure out the delta h and delta s of the reaction as a whole.

00:23 Remember, when we do this, we always do products minus reactives.

00:28 This is the phrase you want to keep repeating to yourself.

00:31 Products minus reactants.

00:33 So for delta h of the reaction, we're going to take the products, so zero here, but we have negative 110 minus the reactants, so minus negative 200, and that's going to give us our delta h, which is 90 .7.

00:55 I didn't include the decimals on this when i was writing it just for the sake of space, but i did use decimals that i found from the appendix and calculation.

01:08 Now, delta s, again, we're going to take products minus reactants.

01:11 So we have 130 .6.

01:13 Remember, there's this coefficient of 2, so it's really 2 times 130 .6 plus 197 for the carbon monoxide here, minus the reactants, products minus reactants, remember that.

01:29 And so this gives us a value for delta s, which is going to be 221.

01:36 Now remember, units, delta s are reported as joules per mole kelvin.

01:42 So that's important on us.

01:45 So now our next step, we were asked to find delta g at various temperatures.

01:50 Remember, delta g is related to delta h minus t delta s with t and kelvin.

01:59 Now the values we're given for the temperature are, i'll say t1 is equal to 28 degrees celsius.

02:12 T2 is 128 and a t3 at 228 degrees celsius.

02:20 And so we need to, for each of these, added 273 to it because we want to work in kelvin.

02:28 And so i'll erase these.

02:31 What we really have as our temperatures that we're going to be plugging in is 301, 401, and 501.

02:43 And so we go on whenever we plug this into this equation, we need to be careful.

02:53 I know that this is in joules, but our delta h is in kilojoules.

02:56 And so what i commonly do is i just turn kilojoules into joules by moving this over by three places...

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