Where possible, classify these systems as reactant-favored or product-favored at 298 K. If the direction cannot be determined from the information given, classify the reaction as "Insufficient information".
a)2A(g) + 3B(g) -->4C(g) delta H = -95kJ
b)2A(g) + 2B(g) -->3C(g) delta H = 254kJ
c)2A(g) + B(g) --> 4C(g) delta H =322 kJ
d)2A(g) + B(g) --> 3C(g) delta H -193kJ
I thought that a)reactant favored, b) reactant favored, c) product favored, and d) insufficient information because that in a and b there are more reactant than products and in c there are more products than reactant and d is equal amount. But I got the answer wrong. Could someone explain to me?
I think D and B have definite answers.
D. 3 mols left; 3 right. dS zero so TdS zero. that + a - dH gives dG - so we go with products favored.
B. 4 mols left; 3 right. dS is - which makes -TdS + and that added to + dH means +dG so reactants favored.
A and B are up for grabs. Here is what I would do.
A. 5 mols left; 4 right so dS is - which makes -TdS +. So the question is can the -dH given overcome the + TdS term? If we guess that dS might be 50 and that x 298, then TdS might be in the neighborhood of 15,000 J. Even at 100 for dS, then TdS is in the neighborhood of 30,000. When added to -95,000 it still carries a - sign and products are favored. All of that depends upon how reasonable we are at guessing for the value of dS. Even at 300 for dS it still favors the products. So I would go with favored; however, since you don't really know you may feel better choosing that you can't tell.
C. 3 mols left; 4 right so dS is + which makes TdS -. Add that to a large (322,000 J) means the TdS term has a lot to overcome to get a - number out of it. Go through the same reasoning as in A above. Even at 400 for dS, then TdS is about 120,000 and that isn't even close to making it -. I would go with reactants favored. I don't have enough experience in thermodynamics to know what reasonable values are for dH and dS.
a insufficient
b reverse
c insufficient
d product
Sure, I'd be happy to explain!
In order to determine whether a reaction is reactant-favored or product-favored, we need to consider the sign of the heat change (delta H) and the stoichiometry of the reaction. Here's a breakdown of each reaction:
a) 2A(g) + 3B(g) --> 4C(g) delta H = -95 kJ
In this reaction, the heat change is negative (exothermic), indicating that the reaction is product-favored. Additionally, there are more reactant molecules (5) than product molecules (4), which further supports that the reaction is reactant-favored.
b) 2A(g) + 2B(g) --> 3C(g) delta H = 254 kJ
Here, the heat change is positive (endothermic), suggesting that the reaction is reactant-favored. However, we can break down the stoichiometry to see that there are actually more product molecules (3) than reactant molecules (4). Based on this, we can conclude that the reaction is product-favored.
c) 2A(g) + B(g) --> 4C(g) delta H = 322 kJ
The heat change in this reaction is positive (endothermic), which indicates that the reaction is reactant-favored. The stoichiometry shows that there are more product molecules (4) than reactant molecules (3), further confirming that the reaction is product-favored.
d) 2A(g) + B(g) --> 3C(g) delta H = -193 kJ
This is where you made a mistake in your initial classification. The heat change is negative (exothermic), suggesting that the reaction is product-favored. However, the stoichiometry shows that there are less product molecules (3) than reactant molecules (3). Based on this, we cannot definitively determine whether the reaction is reactant-favored or product-favored. Therefore, the correct classification is "Insufficient information."
To recap:
a) Product-favored
b) Product-favored
c) Product-favored
d) Insufficient information
I hope this helps clarify the reasoning behind the classifications!
To determine whether a reaction is reactant-favored or product-favored, we can analyze the sign of the enthalpy change (ΔH) for each reaction.
If ΔH is negative (exothermic), the reaction is product-favored, as the formation of products releases energy. If ΔH is positive (endothermic), the reaction is reactant-favored, as the formation of products requires an input of energy.
Now, let's re-evaluate each reaction:
a) 2A(g) + 3B(g) --> 4C(g) ΔH = -95 kJ
The negative ΔH value indicates that the reaction is exothermic and releases energy. Thus, it is a product-favored reaction.
b) 2A(g) + 2B(g) --> 3C(g) ΔH = 254 kJ
The positive ΔH value suggests that the reaction is endothermic and requires an input of energy. This indicates that the reaction is reactant-favored.
c) 2A(g) + B(g) --> 4C(g) ΔH = 322 kJ
Similar to reaction (b), the positive ΔH value implies that the reaction is endothermic and requires an input of energy. Thus, it is reactant-favored.
d) 2A(g) + B(g) --> 3C(g) ΔH = -193 kJ
The negative ΔH value indicates that the reaction is exothermic and releases energy. Therefore, it is a product-favored reaction.
In summary:
a) Product-favored
b) Reactant-favored
c) Reactant-favored
d) Product-favored
From your analysis, you correctly identified the favorability of reactions (a), (c), and (d). However, you mistakenly identified reaction (b) as reactant-favored, when it is actually reactant-favored.
To determine whether a reaction is reactant-favored or product-favored, we can use the concept of Gibbs free energy (ΔG). The equation for Gibbs free energy is given by:
ΔG = ΔH - TΔS
Where:
- ΔG is the change in Gibbs free energy
- ΔH is the change in enthalpy (heat)
- T is the temperature in Kelvin
- ΔS is the change in entropy (disorder)
At constant temperature and pressure, ΔG can be used to determine whether a reaction is spontaneous or not. If ΔG is negative, the reaction is product-favored and spontaneous. If ΔG is positive, the reaction is reactant-favored and non-spontaneous. If ΔG is zero, the reaction is at equilibrium.
Now, let's analyze each reaction:
a) 2A(g) + 3B(g) --> 4C(g) ΔH = -95 kJ
Since ΔH is negative, we can say that the reaction is exothermic. However, we cannot determine the spontaneity of the reaction solely from ΔH. We need additional information about the entropy change (ΔS) to calculate ΔG. Therefore, the direction cannot be determined from the given information. The correct classification is "Insufficient information."
b) 2A(g) + 2B(g) --> 3C(g) ΔH = 254 kJ
In this case, ΔH is positive, indicating an endothermic reaction. Like in the previous example, without the entropy change information (ΔS), we cannot determine the spontaneity of this reaction. Therefore, the correct classification is again "Insufficient information."
c) 2A(g) + B(g) --> 4C(g) ΔH = 322 kJ
Here, we also don't have information about ΔS. However, we can still make an estimation. The reaction shows an increase in the number of moles of gas molecules on the product side (two moles of reactants versus four moles of products). Generally, an increase in the number of gas moles (Δn) favors the product side. Therefore, based on the assumption that ΔS is positive due to the increase in gas molecules, we can classify this reaction as "Product-favored."
d) 2A(g) + B(g) --> 3C(g) ΔH = -193 kJ
Similar to the previous cases, we lack information about ΔS. However, we can still make an estimation. The reaction shows a decrease in the number of moles of gas molecules on the product side (one mole of reactant versus three moles of products). Generally, a decrease in Δn favors the reactant side. Therefore, based on the assumption that ΔS is negative due to the decrease in gas molecules, we can classify this reaction as "Reactant-favored."
To recap:
a) Insufficient information
b) Insufficient information
c) Product-favored
d) Reactant-favored