Extra Credit 21

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Q16.4.14

Calculate the equilibrium constant at the temperature given.

Used the Van't Hoff equation for all of these. (Found K at 25 degrees and ΔH° online, plugged in to equation and solved for K₂)

Van't Hoff Equation:

ln(K₂/K₁)= - ΔH° /R (1/T₂-1/T₁)

(a) \(\ce{I2}(s)+\ce{Cl2}(g)⟶\ce{2ICl}(g) \hspace{20px} \mathrm{(T=100\:°C)}\)
K=0.02765
\(\ce{H2}(g)+\ce{I2}(s)⟶\ce{2HI}(g) \hspace{20px} \mathrm{(T=0.0\:°C)}\)
K=6486.4
\(\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g) \hspace{20px} \mathrm{(T=125\:°C)}\)
Could not find the necessary information required for this problem.
\(\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g) \hspace{20px} \mathrm{(T=675\:°C)}\)
K=11.73
\(\ce{CS2}(g)⟶\ce{CS2}(l) \hspace{20px} \mathrm{(T=90\:°C)}\)
K=0.342

Answer:

a. k=18321

b. k=0.318

c. k=0.318

d. k=1.81 x 10²⁰

e. k=1.716

Q15.1.X

Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Table E3 for solubility products).

TlCl

Ksp=1.86 * 10^-4=[Tl⁺][Cl^-]

[Tl⁺]=1.36 *10^-2

[Cl^-] = 1.36 *10^-2

b.BaF₂

Ksp=1.84 * 10^-7=[Ba²⁺][F^-]²

[Ba²⁺]=x

[F^-] = 2x

[Ba²⁺]=3.58 * 10^-3

[F^-]=7.16 * 10^-3

c.Ag₂CrO₄

Ksp=1.12 * 10^-12=[Ag⁺]²[CrO₄^2-]

[Ag⁺]=2x

[CrO₄^2-] = x

[CrO₄^2-]=6.54 * 10^-5

[Ag⁺]=1.308 * 10^-4

d.CaC₂O₄•H₂O

Ksp=1.96* 10^-8

Ksp=[Ca²⁺][C₂O₄^2-]

[Ca²⁺]=1.4 *10^-4

[C₂O₄^2-]=1.4 *10^-4

e.the mineral anglesite, PbSO₄

Ksp=2.53* 10^-8

Ksp=[Pb²⁺][SO₄^2-]

[Pb²⁺]=1.59 *10^-4

[SO₄^2-]=1.59 *10^-4

Answer:

a. [Tl⁺] = [Cl^-] = 1.36 x 10^-2

b.[Ba²⁺]=3.58 * 10^-3 [F^-]=7.16 x 10^-3

c. [Ag⁺]=1.308 * 10^-4[CrO₄^2-]=6.54 x 10^-5

d. [Ca²⁺] = [C₂O₄^2-]=1.4 x 10^-4

e. [Pb²⁺] = [SO₄^2-]=1.59 x 10^-4

Q14.6.6

What is [OH⁻] in a solution of 0.125 M CH₃NH₂ and 0.130 M CH₃NH₃Cl?

\(\ce{CH3NH2}(aq)+\ce{H2O}(l)⇌\ce{CH3NH3+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=4.4×10^{−4}\)

	CH₃NH₂	CH₃NH₃⁺	OH^-
Initial	0.125	0.13	~0
Change	-x	+x	+x
Equilibrium	0.125-x	0.13+x	x

[(x)(0.13+x)]/(0.125-x) = 4.4 * 10^-4

[OH^-]=x=4.2 * 10^-4 M

Answer:

[OH⁻] in a solution of 0.125 M CH₃NH₂ and 0.130 M CH₃NH₃Cl is 4.2 x 10^-4 M

Q13.3.14

Ammonia is a weak base that reacts with water according to this equation:

\[\ce{NH3}(aq)+\ce{H2O}(l)\rightleftharpoons\ce{NH4+}(aq)+\ce{OH-}(aq)\]

Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water and why?

Addition of NaOH
Addition of HCl
Addition of NH₄Cl

b) Addition of HCl. Due to Le Chatelier's Principle, adding HCl will decrease the concentration of hydroxide, which will then cause the equilibrium to shift to the right, generating more ammonium.

Q5.4.42

How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions?

4Cr + 3O₂ --> 2Cr₂O₃

ΔH= -1128.4 kJ/mol

1.25g Cr * (1 mol Cr/52g) * (2 mol Cr₂O₃/4 mol Cr) * (-1128.4 kJ/mol Cr₂O₃) = -13.5625 kJ

Ethylene, C₂H₂, a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About 80% of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst.

\(\ce{C2H4}(g)+\ce{H2O}(g)⟶\ce{C2H5OH}(l)\)

Using the data in the table in Appendix G, calculate ΔH° for the reaction.

ΔH° = ΔH° (products) - ΔH° (reactants)

(-234.8)-(52.4-285.83)

Answer:

ΔH° for the reaction is -1.37kJ/mol