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
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.
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 +
You're right for #3. dG = 0 at equilibrium.
To determine if a reaction is spontaneous or nonspontaneous, you need to consider the changes in Gibbs free energy (ΔG), enthalpy (ΔH), and entropy (ΔS).
For question #1:
The reaction is given as CaC2(s) + 2H2O(l) ---> Ca(OH)2(s) + C2H2(g).
To determine if the reaction is spontaneous at all temperatures, you need to look at the sign of ΔG. If ΔG is negative, the reaction is spontaneous; if ΔG is positive, the reaction is nonspontaneous; and if ΔG is zero, the reaction is at equilibrium.
In this case, since the reaction is said to be spontaneous at all temperatures, you can conclude that the answer is option E) ΔH is positive and ΔS is negative. This is because if ΔG is negative at all temperatures, then ΔH must be positive and ΔS must be negative based on the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin.
For question #2:
The reaction is given as C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g).
To predict the signs of ΔH and ΔS for the reaction, you need to consider the stoichiometry of the reaction. If the number of gas molecules decreases, ΔS is negative; if the number of gas molecules increases, ΔS is positive.
In this case, the reaction involves the combustion of propane (C3H8) and oxygen (O2), forming carbon dioxide (CO2) and water (H2O). The reaction produces more gas molecules than it consumes, so ΔS is positive. To determine the sign of ΔH, you can consider that combustion reactions usually release energy, making them exothermic, so ΔH is negative.
Therefore, the answer to question #2 is option C) ΔH is negative and ΔS is positive for the reaction.
For question #3:
For a reaction system at equilibrium, the changes in Gibbs free energy, enthalpy, internal energy, and entropy are all zero.
Therefore, the answer to question #3 is option A) ΔG = 0.