Explanation The required reaction is given below. N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 N H 3 ( g ) , K = 0.036 The expression for equilibrium constant for the above reaction can be written as shown below, K = ( P N H 3 ) 2 ( P N 2 ) ( P H 2 ) 3 Given that, the initial partial pressures of N 2 a n d H 2 are 1.42 b a r a n d 2.87 b a r respectively. An equilibrium table can be set up as given below. C o m p o s i t i o n ( b a r ) N 2 H 2 N H 3 I n i t i a l p r e s s u r e 1.42 2.87 0 C h a n g e i n p r e s s u r e − x − 3 x + 2 x E q u i l i b r i u m p r e s s u r e 1.42 − x 2.87 − 3 x 2 x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below. K = ( 2 x ) 2 ( 1.42 − x ) ( 2.87 − 3 x ) 3 By assuming x ≪ 1.42 a n d 2.87 , it can be written that 1.42 − x ≃ 1

Chemical Principles: The Quest for Insight

7th Edition

ISBN: 9781464183959

Author: Peter Atkins, Loretta Jones, Leroy Laverman

Publisher: W. H. Freeman

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Chapter 5, Problem 5I.22E

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Interpretation:

The equilibrium partial pressure of each substance in the mixture has to be calculated. The reaction is given below.

N2(g)+3H2(g)⇌2NH3(g), K=0.036

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Chapter 5, Problem 5I.21E

Chapter 5, Problem 5I.23E

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In the Harber process for ammonia synthesis, K= 0.036 for N2(g) + 3H2(g) ⇋ 2 NH3(g) at 500. K. If the reactor of vlume 2.0 L charged with 1.42 bar N2 and 2.87 bar H2, what will the equilibrium partial pressures in the mixture?

Carbon monoxide and water vapor, each at 200. Torr, were introduced into a container of volume 0.250 L. When the mixture reached equilibrium at 700 degrees Celsius, the partial pressure of CO2(g) was 88 Torr. Calculate the value of K for the equilibrium CO (g) + H2O (g)⇋ CO2(g)+H2(g).

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The equilibrium partial pressure of each substance in the mixture has to be calculated. The reaction is given below. N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 N H 3 ( g ) , K = 0.036

Solution Summary: The author explains how the equilibrium partial pressure of each substance in the mixture has to be calculated.

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Interpretation:

The equilibrium partial pressure of each substance in the mixture has to be calculated. The reaction is given below.

N2(g)+3H2(g)⇌2NH3(g), K=0.036

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Chapter 5 Solutions

The equilibrium partial pressure of each substance in the mixture has to be calculated. The reaction is given below. N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 N H 3 ( g ) , K = 0.036

Solution Summary: The author explains how the equilibrium partial pressure of each substance in the mixture has to be calculated.

Videos

Interpretation: The equilibrium partial pressure of each substance in the mixture has to be calculated. The reaction is given below. N2(g)+3H2(g)⇌2NH3(g), K=0.036

Want to see the full answer?

Chapter 5 Solutions

Interpretation:

The equilibrium partial pressure of each substance in the mixture has to be calculated. The reaction is given below.

N2(g)+3H2(g)⇌2NH3(g), K=0.036