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Orlov [11]
3 years ago
13

Carbon monoxide and chlorine gas react to produce COCl2 gas. The Kp for the reaction is 1.49 × 108 at 100.0 °C: In an equilibriu

m mixture of the three gases, PCO, = PCl2 = 7.70 × 10-4 atm. The partial pressure of the product, phosgene (COCl2), is ________ atm. Use the assumption that change in x is small so you do not have to use the quadratic.
Chemistry
2 answers:
Eva8 [605]3 years ago
5 0

Answer:

Explanation:

Solution:

For the equilibrium

The equilibrium constant is defined in terms of partial pressure:

Introducing the numerical data given for partial pressureof carbon monoxide 0 and chlorine 12, also the value for equilibrium constant:

Answer:

The partial pressureof the product, phosgene (COCl2), is 29.4atm

liberstina [14]3 years ago
3 0

Answer:

88.34 atm

Explanation:

At equilibrium, carbon monoxide (CO) would react with chlorine gas according to the equation below:

CO (g) + Cl₂ (g)   ⇒ COCl₂ (g)

The equilibrium constant Kp, which is a ratio of the partial pressure of the product to that of the reactants is obtained from the equation below:

Kp = PCOCL / PCO.PCl₂

From the question given,

Kp = 1.49 x 10⁸ t 100° C

PCO = 7.70 X 10⁻⁴ atm

PCl₂ = 7.70 x 10⁻⁴ atm

It therefore implies that,

1.49 x 10⁸  = P(COCl₂) / (7.70 x 10⁻⁴). (7.70 x 10⁻⁴)

P(COCl₂) = 1.49 x 10⁸) . (7.70 x 10⁻⁴) . (7.70 x 10⁻⁴)

P (COCl₂) = 88.34 atm

The partial pressure P(COCl₂) of the product phosgene (COCl₂) is 88.34 atm

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How many grams of fluorine are contained in 8 molecules of boron trifluoride?
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<h3>Answer:</h3>

             7.57 × 10⁻²² g of F

<h3>Solution:</h3>

Data Given:

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                 M.Mass of BF₃ =  67.82 g.mol⁻¹

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Step 1: Calculate Moles of BF₃

           Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Putting value,

            Moles  =   8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

            Moles  =  1.33 × 10⁻²³ mol

Step 2: Calculate Mass of BF₃:

                   Moles  =  Mass ÷ M.Mass

Solving for Mass,

                   Mass  =  Moles × M.Mass

Putting values,

                   Mass  =  1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

                   Mass  =  9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

                         67.82 g BF₃ contains  =  57 g of F

So,

                    9.0 × 10⁻²² g will contain  =  X g of F

Solving for X,

                       X =  (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

                        X  =  7.57 × 10⁻²² g of F

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Answer:

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