1. The following reaction is spontaneous at all temperatures:
CaC2(s) + 2H2O(l) ---> Ca(OH)2(s) + C2H2(g)
Which of the following statements is true?
A) ΔG is positive at all temperatures.
B) ΔG is positive and ΔS is positive.
C) ΔH is negative and ΔS is negative.
D) ΔH is negative and ΔS is positive.
E) ΔH is positive and ΔS is negative.
Ok. We are working with
dG = dH -TdS. To be spontaneous dG must be negative so that means A and B can't right.
For C, dG = dH -TdS. If dH is - (that gives us the right start to get a - number) and if dS is - then that -TdS term will always be + so the only way dG can be - is for dH to be a larger - term than the -TdS term so C can't be right.
D. If dH is - (again a good start) and dS is + (then the term - TdS will always be negative) and if we add a -dH to a - TdS we ALWAYS get a negative number so dG will be negative and the rxn is spontaneous. So I think #1 is D.
E, which you picked, can't be right at ANY temperature. If dH is + (a bad start) and dS is - then the terms -TdS will always be + and if we add + to + we get a + so dG can't be spontaneous at any temperature.
I would really like to understand how to do these problems, because my exam is coming up in a few weeks. I still cannot comprehend how you can tell if a system is is positive or negative according to reaction.
For #1, I presume it is E if it is spontaneous at all temperatures. For #2, I think ΔS have to be negative since the reaction contains all gas phases. But I have no idea if you can tell if ΔH is spontaneous or nonspontaneous..And for #3, I presume it is A(?) basing on Gibbs Free Energy.
1. The following reaction is spontaneous at all temperatures:
CaC2(s) + 2H2O(l) ---> Ca(OH)2(s) + C2H2(g)
Which of the following statements is true?
A) ΔG is positive at all temperatures.
B) ΔG is positive and ΔS is positive.
C) ΔH is negative and ΔS is negative.
D) ΔH is negative and ΔS is positive.
E) ΔH is positive and ΔS is negative.
2. Consider the following reaction:
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g)
One would predict that
A) ΔH is positive and ΔS is positive for the reaction.
B) ΔH is negative and ΔS is negative for the reaction.
C) ΔH is negative and ΔS is positive for the reaction.
D) ΔH is positive and ΔS is negative for the reaction.
E) ΔG is positive at all temperatures.
3. For the reaction system that is at equilibrium, which of the following must always be true?
A) ΔG = 0
B) ΔH = 0
C) ΔU = 0
D) ΔS = 0
E) q = 0
3 answers
2. Consider the following reaction:
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g)
One would predict that
A) ΔH is positive and ΔS is positive for the reaction.
B) ΔH is negative and ΔS is negative for the reaction.
C) ΔH is negative and ΔS is positive for the reaction.
D) ΔH is positive and ΔS is negative for the reaction.
E) ΔG is positive at all temperatures.
You're right about dS. That is + based on 7 moles gas on the right and 6 on the left. There are two ways to do the dH value.
1. Look at bonds formed vs bonds broken. If B>F dH is -; if B<F dH is +. But that can get complicated, especially if double or triple bonds are involved since a double or triple bond is not 2x or 3x a single bond. You may also use Hess' law and look up dHf and calculate dHf rxn.
I believe the answer is C. dH is - and dS is +
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g)
One would predict that
A) ΔH is positive and ΔS is positive for the reaction.
B) ΔH is negative and ΔS is negative for the reaction.
C) ΔH is negative and ΔS is positive for the reaction.
D) ΔH is positive and ΔS is negative for the reaction.
E) ΔG is positive at all temperatures.
You're right about dS. That is + based on 7 moles gas on the right and 6 on the left. There are two ways to do the dH value.
1. Look at bonds formed vs bonds broken. If B>F dH is -; if B<F dH is +. But that can get complicated, especially if double or triple bonds are involved since a double or triple bond is not 2x or 3x a single bond. You may also use Hess' law and look up dHf and calculate dHf rxn.
I believe the answer is C. dH is - and dS is +
You're right for #3. dG = 0 at equilibrium.