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:
[NO] = 1.72 x 10⁻³ M.
Explanation:
<em>2NO(g) ⇌ N₂(g)+O₂(g),</em>
Kc = [N₂][O₂] / [NO]².
- At initial time: [NO] = 0.171 M, [N₂] = [O₂] = 0.0 M.
- At equilibrium: [NO] = 0.171 M - 2x , [N₂] = [O₂] = x M.
∵ Kc = [N₂][O₂] / [NO]².
∴ 2400 = x² / (0.171 - 2x)² .
<u><em>Taking the aquare root for both sides:</em></u>
√(2400) = x / (0.171 - 2x)
48.99 = x / (0.171 - 2x)
48.99 (0.171 - 2x) = x
8.377 - 97.98 x = x
8.377 = 98.98 x.
∴ x = 8.464 x 10⁻².
<em>∴ [NO] = 0.171 - 2(8.464 x 10⁻²) = 1.72 x 10⁻³ M. </em>
<em>∴ [N₂] = [O₂] = x = 8.464 x 10⁻² M.</em>
Answer:
The plant life cycle starts when a seed falls on the ground. There are many different kinds of plant life, but the flowering plants, or angiosperms, are the most advanced and widespread due to their amazing ability to attract pollinators and spread seeds.
Answer:
133.74 L
Explanation:
First we <u>convert the given pressures and temperatures into atm and K</u>, respectively:
- 750.0 Torr ⇒ 750/760 = 0.9868 atm
- 20°C ⇒ 20+273.16 = 293.16 K
- 40°C ⇒ 40+273.16 = 313.16 K
Then we<u> use the PV=nRT formula to calculate the number of moles of helium in the balloon</u>, using<em> the data of when it was on the ground</em>:
- 0.9868 atm * 8.50 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293.16 K
Then, knowing the value of n, we <u>use PV=nRT once again, this time to calculate V</u> using <em>the data of when the balloon was high up:</em>
- 0.550 atm * V = 2.866 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 313.16 K
Shown above are three simple diatomic molecules. On the left is H2, with a single bond between the H atoms. Since two H atoms have the same electronegativity, they share the electrons evenly. Fluorine is much more electronegative than hydrogen, so the majority of electron density resides on the fluorine atom.
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