Two moles HI were heated in a sealed tube at 440C till the equilibrium was reached.HIwas found to be 22% decomposed.The equilibrium constant for dissociation is?

Question

pratham ashish · Answer

Hi, ther e r 2of eq. constant K c & K p rxn 2HI -> H 2 +I 2 moles initally 2 0 0 moles @eqlb. 2-2x x x K c = x*x/{2-2x}^2 = (0.22)^2/4*(0.78)^2 = 0.19888 conc.(C)= P/RT for gas phase rxn RT *( x)= P H2 = P I2 = 0.19888 (same as K c bcz power o RT cancel out )

devesh · Answer

there are 2 equilibrium constant K c and K p The reaction, 2 H I ⟶ H 2 ​ + I 2 ​ moles initially 2 0 0 moles of equilibrium 2 − 2 x x x K c = ( 2 − 2 x ) 2 x × x ​ = ( 2 − 2 x ) 2 x × x ​ = 4 × ( 0 . 7 8 ) 2 ( 0 . 2 2 ) 2 ​ =Concentration(C) = R T P ​ = for gas phase reaction R T × X = P H 2 ​ ​ = P I 2 ​ ​ = 0 . 1 9 8 8 8

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Two moles HI were heated in a sealed tube at 440C till the equilibrium was reached.HIwas found to be 22% decomposed.The equilibrium constant for dissociation is?

Two moles HI were heated in a sealed tube at 440C till the equilibrium was reached.HIwas found to be 22% decomposed.The equilibrium constant for dissociation is?

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Two moles HI were heated in a sealed tube at 440C till the equilibrium was reached.HIwas found to be 22% decomposed.The equilibrium constant for dissociation is?

Two moles HI were heated in a sealed tube at 440C till the equilibrium was reached.HIwas found to be 22% decomposed.The equilibrium constant for dissociation is?

2 Answers

there are 2 equilibrium constant Kc and KpThe reaction, 2HI⟶H2​+I2​moles initially 2 0 0moles of equilibrium 2−2x x xKc=(2−2x)2x×x​=(2−2x)2x×x​=4×(0.78)2(0.22)2​=Concentration(C) =RTP​= for gas phase reactionRT×X=PH2​​=PI2​​=0.19888

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there are 2 equilibrium constant Kc and Kp
The reaction, 2HI⟶H2+I2
moles initially 2 0 0
moles of equilibrium 2−2x x x
Kc=(2−2x)2x×x
=(2−2x)2x×x=4×(0.78)2(0.22)2
=Concentration(C) =RTP= for gas phase reaction
RT×X=PH2=PI2=0.19888