Explanation Given The equilibrium constant ( K p ) for the given reaction is 2.26 × 10 − 4 . The initial partial pressure of methane ( CH 4 ) is 105.1 torr . The initial partial pressure of I 2 is 7.96 torr . The given reaction is, CH 4 ( g ) + I 2 ( g ) ⇌ CH 3 I ( g ) + HI ( g ) According to the above equation, the partial pressure of CH 4 and I 2 decreases while the partial pressure of CH 3 I and HI increases. Make an ICE table for the above reaction. CH 4 ( g ) + I 2 ( g ) ⇌ CH 3 I ( g ) + HI ( g ) Initial ( torr ) : 105.1 7.96 0.0 0.0 Change ( torr ) : − x − x + x + x Equilibrium ( torr ) : 105.1 − x 7.96 − x x x The equilibrium constant of the reaction is, K p = ( P CH 3 I ) ( P HI ) ( P CH 4 ) ( P I 2 ) Where, K p is the equilibrium constant of the reaction. P CH 3 I is the equilibrium partial pressure of CH 3 I . P HI is the equilibrium partial pressure of HI . P CH 4 is the equilibrium partial pressure of CH 4 . P I 2 is the equilibrium partial pressure of I 2 . Substitute the value of K p , P CH 3 I , P HI , P CH 4 and P I 2 in the above equation

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ISBN: 9781323233252

Author: Brown

Publisher: PEARSON C

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Chapter 2, Problem 37E

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To determine: The equilibrium partial pressure of all the reactants and products.

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Chapter 2, Problem 36E

Chapter 2, Problem 38E

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