Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
Answer:
n = 5 to n = 6 absorption
n = 9 to n = 6 emission
n = 6 to n = 4 emission
n = 6 to n = 7 absorption
Explanation:
According to the Bohr's model of the atom. An electron in an atom may absorb energy and move from a lower energy level to a higher energy level. This requires absorption of energy that is equal to the energy difference between the two levels.
Similarly, an electron may move from a higher to a lower energy level, releasing energy that is equal to the energy difference between the higher and the lower level. This is known as emission.
Hence, if the electron is moving from a lower energy level to a higher energy level, an absorption has taken place, e.g n = 5 to n = 6
When an electron moves from a higher energy level to a lower energy level, an emission has taken place e.g n = 9 to n = 6
They have different number of neutrons.
