Explanation The required reaction is given below. 2 H 2 S ( g ) ⇌ 2 H 2 ( g ) + S 2 ( g ) The expression for equilibrium constant for the above reaction can be written as shown below, K c = [ H 2 ] [ S 2 ] [ H 2 S ] Given that, 0.1 m o l H 2 S is placed in a reaction vessel of volume 10.00 L . The concentration of H 2 S can be calculated as given below. [ H 2 S ] = m o l e s v o l u m e = 0.1 m o l 10.00 L = 0 . 0 1 M An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) H 2 S H 2 S 2 I n i t i a l c o n c e n t r a t i o n 0.01 0 0 C h a n g e i n c o n c e n t r a t i o n − 2 x + 2 x + x E q u i l i b r i u m c o n c e n t r a t i o n 0.01 − 2 x 2 x x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below. K c = ( 2 x ) 2 ( x ) ( 0.01 − x ) 2 By assuming x ≪ 0.01 , it can be written that 0.01 − x ≃ 0.01. Now, K c = ( 2 x ) 2 ( x ) ( 0.01 ) 2 It is given that, at equilibrium the number of moles of hydrogen gas is 0

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Author: ATKINS

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Chapter 5, Problem 5I.26E

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The value of Kc for the reaction 2H2S(g)⇌2H2(g)+S2(g) has to be calculated.

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Chapter 5, Problem 5I.25E

Chapter 5, Problem 5I.27E

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The value of K c for the reaction 2 H 2 S ( g ) ⇌ 2 H 2 ( g ) + S 2 ( g ) has to be calculated.

Solution Summary: The author explains how the value of K_c for the reaction is calculated. The equilibrium constant expression can be written as shown below.

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Interpretation:

The value of Kc for the reaction 2H2S(g)⇌2H2(g)+S2(g) has to be calculated.

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Chapter 5 Solutions

The value of K c for the reaction 2 H 2 S ( g ) ⇌ 2 H 2 ( g ) + S 2 ( g ) has to be calculated.

Solution Summary: The author explains how the value of K_c for the reaction is calculated. The equilibrium constant expression can be written as shown below.

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Interpretation: The value of Kc for the reaction 2H2S(g)⇌2H2(g)+S2(g) has to be calculated.

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Chapter 5 Solutions

Interpretation:

The value of Kc for the reaction 2H2S(g)⇌2H2(g)+S2(g) has to be calculated.