[Corribus]

Right.  So let's assume for the moment you set the problem up right and only made a math mistake, because you clearly made a math mistake.  At this point all you can do is scrutinize it until you figure out what you did wrong.  I'll help you out.  Your "a, b and c" values you used for the quadratic equation: one of them is not right.

Beyond that, here's a tip: if you know for a fact your answer isn't right, always look first to see if you made a math mistake before you question whether you approached the problem incorrectly.  Only once you've ruled out a math error should you start to question your conceptual approach to the problem.

[webassignbuddy]

Right.  So let's assume for the moment you set the problem up right and only made a math mistake, because you clearly made a math mistake.  At this point all you can do is scrutinize it until you figure out what you did wrong.  I'll help you out.  Your "a, b and c" values you used for the quadratic equation: one of them is not right.

Beyond that, here's a tip: if you know for a fact your answer isn't wrong, always look first to see if you made a math mistake before you question whether you approached the problem incorrectly.  Only once you've ruled out a math error should you start to question your conceptual approach to the problem.

Yeah it was my algebra
Ughhhhh. I can't afford to make these careless mistakes on the test.

I used an online quadratic calculator and input all of my A B and C values and it got x = 0.0759, which is the answer...
And c is supposed to be -0.04524
At least the table was right !

[Corribus]

Good.  I hope you see how, lacking the correct answer as a check, you could easily go through and find that you got an answer that didn't make sense, and then troubleshooted to see where your mistake was.  BTW, if you have a graphing calculator, most have a root function for a polynomial which can be helpful to avoid that kind of error.

Now, onto part B: What are your thoughts here?  I suppose we'll verify your answer in part D, so you can go ahead and take a stab at that while you're at it.  (You have the answer to B already, but maybe you can think of WHY the Eq shifts to the right when the temperature is raised.

[webassignbuddy]

Good.  I hope you see how, lacking the correct answer as a check, you could easily go through and find that you got an answer that didn't make sense, and then troubleshooted to see where your mistake was.  BTW, if you have a graphing calculator, most have a root function for a polynomial which can be helpful to avoid that kind of error.

Now, onto part B: What are your thoughts here?  I suppose we'll verify your answer in part D, so you can go ahead and take a stab at that while you're at it.  (You have the answer to B already, but maybe you can think of WHY the Eq shifts to the right when the temperature is raised.

For part B, I was actually gonna ask you But I'll take a guess...

Well if the Kd (which is referred to as the DISSOCIATION CONSTANT) of the RP is HIGHER, than MORE products are formed, which would mean that the product side is favored and thus the equilibrium shifts to the right.

[Corribus]

Seems like a reasonable answer to me.

What about C?

[webassignbuddy]

Seems like a reasonable answer to me.

What about C?

Reaction is endothermic because the INPUT of heat is required to BREAK bonds and for RP to dissociate into R and P (?)

[Corribus]

Input of heat is almost always required to drive a reaction initially, be it endothermic or exothermic.  It's not a bad thought, but there's a way to show it explicitly.  Here's a hint: you're given two equilibrium constants at two temperatures.  Note: you don't actually have to calculate anything, just determine whether the enthalpy change is positive or negative.

[webassignbuddy]

Input of heat is almost always required to drive a reaction initially, be it endothermic or exothermic.  It's not a bad thought, but there's a way to show it explicitly.  Here's a hint: you're given two equilibrium constants at two temperatures.  Note: you don't actually have to calculate anything, just determine whether the entropy change is positive or negative.

The entropy change should be negative, which would mean that Gibbs free energy would be positive (based on the equation ΔG = ΔH - TΔS ), and the reaction would be endothermic.

[Corribus]

Try again:

[tex]\ln \frac{k_2}{k_1}=\frac{-\Delta H^o}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]

[webassignbuddy]

Try again:

[tex]\ln \frac{k_2}{k_1}=\frac{-\Delta H^o}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]

Ohhh, if you plug everything in and solve for ΔH, it would be a positive value, which means that it's endothermic.

[Corribus]

Exactly.  But you don't actually have to plug anything in.  Let K2 be the higher temperature and K1 be the lower temperature. THis means K2/K1 is greater than one, and the log of any number greater than 1 is positive.  Thus the left hand side is positive.  Likewise, the temperature term has to be negative because if T2>T1, then the reciprocal of T2 has to be smaller than the reciprocal of T1.  Just by looking at the signs, then, it's easy to see how the enthalpy change must be positive, or the reaction is endothermic.  In general, if the equilibrium constant increases when the temperature increases, the process is endothermic. There's a way to rationalize this with a reaction coordinate diagram, but I wouldn't worry about it now.  Just remember the Van't Hoff eqn and you can't go wrong.

Now have a shot at D.  Should be pretty easy, since you should solve it exactly like you did A.

[webassignbuddy]

Exactly.  But you don't actually have to plug anything in.  Let K2 be the higher temperature and K1 be the lower temperature. THis means K2/K1 is greater than one, and the log of any number greater than 1 is positive.  Thus the left hand side is positive.  Likewise, the temperature term has to be negative because if T2>T1, then the reciprocal of T2 has to be smaller than the reciprocal of T1.  Just by looking at the signs, then, it's easy to see how the enthalpy change must be positive, or the reaction is endothermic.  In general, if the equilibrium constant increases when the temperature increases, the process is endothermic. There's a way to rationalize this with a reaction coordinate diagram, but I wouldn't worry about it now.  Just remember the Van't Hoff eqn and you can't go wrong.

Now have a shot at D.  Should be pretty easy, since you should solve it exactly like you did A.

I wrote down that Van't Hoff equation in bold letters and with a pen on a flashcard Definitely will not forget that one.

And my teacher told us to skip d since it's pretty much the same as a except with a different K value (1.47).

Thanks a BUNCH for helping me out!!

Imma try to give the last 2 problems a shot and If I need more help I'll most likely come back here... but they seem pretty straightforward

[Corribus]

Sure no problem.