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galben [10]
3 years ago
8

For the reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g) at 600.0 K, the equilibrium constant is 11.5. Suppose that 1.500 g of PCl5 (MW=208.

22 g/mol) is placed in an evacuated 500.0 mL bulb, which is then heated to 600.0 K. What is the total pressure (in atm) in the bulb at equilibrium?
Chemistry
1 answer:
spayn [35]3 years ago
8 0

Answer:

1.418688 atm

Explanation:

(a)  Moles of PCl5 = mass / molar mass

                      =1.5 g / 208.22 g/mol

                     = 0.0072 moles

Also given,

T = 600 K

V = 0.500 L

Pressure of PCl5, P = nRT / V

                             = 0.0072 mol×0.0821 L-atm / (mol.K)×600 K / 0.500 L

                             = 0.709344 atm

(b)  PCl5(g) ⇄ PCl3(g) + Cl2(g)

Initial               0.965         0               0

Change               -x            +x             +x

Equilibrium (0.709344 -x)         x               x

K_p = 11.5 = x×x / (0.965 -x)

solving, we get x = 0.67027

So partial pressure of PCl5 at equilibrium = 0.709344 - 0.67027 = 0.039074 atm

(c)   Partial pressure of PCl3 = Cl2 = 0.709344 atm

So total pressure = 0.709344+0.039074+ 0.67027= 1.418688 atm

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Answer:

See explanation

Explanation:

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<em> </em>

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In an elimination reaction, a geometry where the β hydrogen and the leaving group are on opposite sides of the molecule is called <u>anti</u> periplanar.

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while in an <u>E1</u><u> </u>mechanism, a base removes a β hydrogen from the carbocation, forming a new π-bond.

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6 0
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The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com
Elanso [62]

Answer:

(a) Ksp=4.50x10^{-7}

(b) Ksp=1.55x10^{-6}

(c) Ksp=2.27x10^{-12}

(d) Ksp=1.05x10^{-22}

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)

Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L}   )^2\\\\Ksp=4.50x10^{-7}

(B) Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)

Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}

(C) NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)

Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}

(D) La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)

Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}

Best regards.

7 0
3 years ago
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