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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?
Hi, there r 2of eq. constant Kc & Kp rxn 2HI -> H2 +I2 moles initally 2 0 0 moles @eqlb. 2-2x x x Kc = x*x/{2-2x}^2 = (0.22)^2/4*(0.78)^2 = 0.19888 conc.(C)= P/RT for gas phase rxn RT *(x)= PH2 = PI2 = 0.19888 (same as Kc bcz power o RT cancel out )
Hi,
there r 2of eq. constant
Kc & Kp
rxn 2HI -> H2 +I2
moles initally 2 0 0
moles @eqlb. 2-2x x x
Kc = x*x/{2-2x}^2 = (0.22)^2/4*(0.78)^2 = 0.19888
conc.(C)= P/RT for gas phase rxn
RT *(x)= PH2 = PI2 = 0.19888 (same as Kc bcz power o RT cancel out )
there are 2 equilibrium constant Kc and KpThe reaction, 2HI⟶H2+I2moles 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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