1. Calculate ∆Sº for the reaction below:
N2(g)+O2(g)⇄2NO(g)
where ∆Sº for N2(g), O2(g), & NO(g), respectively, is 191.5, 205.0, & 210.6 J/mol-K
a. -24.7 J/K
b. 24.7 J/K
c. 185.9 J/K ***
d. -185.9 J/K
2. For which of these is there a decrease in entropy?
a. Cl2(g)→2Cl(g)
b. O2(g)→O2(aq)
c. C(s)+O2(g)→CO2(g)
d. NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l) ***it created a liquid
1. 205.0 + 191.5 = 396.5
210.6 - 396.5= -185.9 j/k
You didn't follow instructions.
"where ∆Sº for N2(g), O2(g), & NO(g), respectively, is 191.5, 205.0, & 210.6"
so N2 = 191.5
O2 = 205.0
NO = 210.6
From my previous post:
N2(g)+O2(g)⇄2NO(g)
dSo rxn = (n*dSo products) - (n*dSo reactants)
dSo rxn = (2*dSo NO) - (1*dSo N2 + 1*dSo O2 =
2*210.6 - 191.5 - 205.0 = ?
I change my answer to 24.7...forgot the 2
1. To calculate ∆Sº for a reaction, you need to use the equation:
∆Sº = Σn(∆Sº products) - Σn(∆Sº reactants)
where Σn represents the stoichiometric coefficients of the species involved in the reaction and (∆Sº) represents the standard molar entropy.
In the given reaction, the stoichiometric coefficients are 1 for N2(g), 1 for O2(g), and 2 for NO(g). The standard molar entropies for N2(g), O2(g), and NO(g) are 191.5, 205.0, and 210.6 J/mol-K respectively.
∆Sº = (2 * ∆Sº NO) - (∆Sº N2 + ∆Sº O2)
= (2 * 210.6) - (191.5 + 205.0)
= 421.2 - 396.5
= 24.7 J/K
Therefore, the value of ∆Sº for the given reaction is 24.7 J/K.
The correct answer is b. 24.7 J/K.
2. To determine which of the given reactions results in a decrease in entropy, you need to compare the total entropy of the reactants to the total entropy of the products.
a. Cl2(g)→2Cl(g)
In this reaction, the number of gaseous molecules decreases from 1 to 2. As gas molecules have higher entropy than liquid or solid molecules, there is an increase in entropy. Therefore, there is no decrease in entropy for this reaction.
b. O2(g)→O2(aq)
In this reaction, O2 gas dissolves in water to form an aqueous solution. The number of gas molecules remains the same, but the entropy may decrease due to the formation of a more structured solvated species. So there could be a decrease in entropy.
c. C(s)+O2(g)→CO2(g)
In this reaction, a solid and a gas react to form a gas. The number of gas molecules increases, resulting in an increase in entropy.
d. NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)
In this reaction, two aqueous solutions react to form a new aqueous solution and water. The formation of a liquid (water) from aqueous species usually leads to a decrease in entropy, as liquids have lower entropy than solutions with dissociated ions. Therefore, this reaction results in a decrease in entropy.
Therefore, the correct answer is d. NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l).
1. dSo rxn = (n*dSo products) - (n*dSo reactants)
2. You want the process in which movement of the molecules is decreased.