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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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3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
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