Chpt%2017-10

Chpt%2017-10 - Equilibrium Calculations 17.3 17.5 Examples...

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Equilibrium Calculations 17.3 – 17.5
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Examples Given: N 2 ( g ) + O 2 ( g ) 2 NO( g ) K = 4.3 * 10 -25 Find K for: 3 N 2 ( g ) + 3 O 2 ( g ) 6 NO( g ) 4 NO( g ) 2 N 2 ( g ) + 2 O 2 ( g )
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Modifying K (and Q) Combining reactions: N 2 ( g ) + 3 H 2 ( g ) 2NH 3 ( g ) K = 0.0024 + 2NH 3 ( g ) + 2H 2 O( l ) 2NH 4 + ( aq ) + 2OH - ( aq ) K = 0.000075 ------------------------------------------ N 2 ( g ) + 3H 2 ( g ) + 2H 2 O( l ) 2NH 4 + ( aq ) + 2OH - ( aq ) K = 0.0024 * 0.000075 = 1.8 * 10 -7
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17.3 K c and K p Reactions in Solution Denoted by ( aq ) Concentration measured by Molarity The K value is referred to as K c Reactions in Gas Denoted by ( g ) Concentration measured by atm The K value is referred to as K p There is a relationship between K c and K p We will cover that later… right now, everything is K c
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17.5 Equil and Stoich Consider a reaction: A( aq ) + B( aq ) C( aq ) K c = 3 Given initial conditions of [A] = 1 M and [B] = 1 M and [C]
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Chpt%2017-10 - Equilibrium Calculations 17.3 17.5 Examples...

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