1.

2 mol of HI was heated in a sealed tube at 440°C till equilibrium was reached. HI was found tobe 22% decomposed. The equilibrium constant for the dissociation is(A) 0.282(B) 0.0796(C) 0.0199(D) 1.99

Answer»

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 Kcbcz power o RT cancel out )



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