Explanation The required reaction is given below. N 2 ( g ) + O 2 ( g ) ⇌ 2 N O ( g ) , K = 3.4 × 10 − 21 The expression for equilibrium constant for the above reaction can be written as shown below, K = ( P N O ) 2 ( P N 2 ) ( P O 2 ) Given that, the equilibrium partial pressures of N 2 a n d O 2 are both 52 k P a . by rearranging the above expression, the equilibrium partial pressure of N O can be calculated as shown below. K = ( P N O ) 2 ( P N 2 ) ( P O 2 ) ( P N O ) 2 = K × ( P N 2 ) ( P O 2 ) ( P N O ) 2 = ( 3

7th Edition

ISBN: 9781319086411

Author: ATKINS

Publisher: MAC HIGHER

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Question

Chapter 5, Problem 5I.1BST

Interpretation Introduction

Interpretation:

The equilibrium partial pressure of NO at 800 K for the given below reaction has to be calculated.

N2(g)+O2(g)⇌ 2NO(g), K=3.4×10−21

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Chapter 5, Problem 5I.1AST

Chapter 5, Problem 5I.2AST

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Chapter 5 Solutions

CHEMICAL PRINCIPLES PKG W/SAPLING

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Knowledge Booster

The equilibrium partial pressure of N O at 800 K for the given below reaction has to be calculated. N 2 ( g ) + O 2 ( g ) ⇌ 2 N O ( g ) , K = 3.4 × 10 − 21

Solution Summary: The author explains the equilibrium partial pressure of NO at 800K for the given below reaction has to be calculated.

Interpretation:

The equilibrium partial pressure of NO at 800 K for the given below reaction has to be calculated.

N2(g)+O2(g)⇌ 2NO(g), K=3.4×10−21

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Chapter 5 Solutions

The equilibrium partial pressure of N O at 800 K for the given below reaction has to be calculated. N 2 ( g ) + O 2 ( g ) ⇌ 2 N O ( g ) , K = 3.4 × 10 − 21

Solution Summary: The author explains the equilibrium partial pressure of NO at 800K for the given below reaction has to be calculated.

Interpretation: The equilibrium partial pressure of NO at 800 K for the given below reaction has to be calculated. N2(g)+O2(g)⇌ 2NO(g), K=3.4×10−21

Want to see the full answer?

Chapter 5 Solutions

Interpretation:

The equilibrium partial pressure of NO at 800 K for the given below reaction has to be calculated.

N2(g)+O2(g)⇌ 2NO(g), K=3.4×10−21