Explanation The required reaction is given below. N 2 ( g ) + O 2 ( s ) ⇌ 2 N O ( g ) , K c = 1.00 × 10 − 5 The expression for equilibrium constant for the above reaction can be written as shown below, K c = [ N O ] 2 [ N 2 ] [ O 2 ] Given that, the number of moles of N 2 a n d O 2 initially present in the reaction vessel of volume 10.00 L is 0.312 m o l a n d 0.407 m o l respectively. Their concentrations can be calculated as given below. [ N 2 ] = m o l e s v o l u m e = 0.312 m o l 10.00 L = 0 . 0 3 1 2 M [ O 2 ] = m o l e s v o l u m e = 0.4 07 m o l 10.00 L = 0 . 0 4 0 7 M An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) N 2 O 2 N O I n i t i a l c o n c e n t r a t i o n 0.0312 0.0407 0 C h a n g e i n c o n c e n t r a t i o n − x − x + 2 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 0.0312 − x 0.0407 − x 2 x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below. K c = ( 2 x ) 2 ( 0.0312 − x ) ( 0.0407 − x ) By assuming x ≪ 0

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Chapter 5, Problem 5I.18E

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Interpretation:

The equilibrium molar concentrations of NO, N2, and O2 in a reaction vessel of volume 10.00 L that initially held 0.312 mol N2 and 0.407 mol O2 have to be calculated. The reaction is given below.

N2(g)+O2(s)⇌2NO(g), Kc=1.00×10−5

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Chapter 5, Problem 5I.17E

Chapter 5, Problem 5I.19E

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Chapter 5 Solutions

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The equilibrium molar concentrations of N O , N 2 , a n d O 2 in a reaction vessel of volume 10.00 L that initially held 0.312 m o l N 2 a n d 0.407 m o l O 2 have to be calculated. The reaction is given below. N 2 ( g ) + O 2 ( s ) ⇌ 2 N O ( g ) , K c = 1.00 × 10 − 5

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Interpretation: The equilibrium molar concentrations of NO, N2, and O2 in a reaction vessel of volume 10.00 L that initially held 0.312 mol N2 and 0.407 mol O2 have to be calculated. The reaction is given below. N2(g)+O2(s)⇌2NO(g), Kc=1.00×10−5

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Chapter 5 Solutions

Interpretation:

The equilibrium molar concentrations of NO, N2, and O2 in a reaction vessel of volume 10.00 L that initially held 0.312 mol N2 and 0.407 mol O2 have to be calculated. The reaction is given below.

N2(g)+O2(s)⇌2NO(g), Kc=1.00×10−5