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`I^(-)` or the reaction `I_(2) (g) iff 2I(g)`. `K_(c )=6.25xx10^(-6)` at `900^(@)C`. If 1 mole of `I_(2)` injected into a 4 litre of box at `900^(@)C` initially. Then select correct statement(s). [Given : R = 0.08 L-atom `K^(-1) mol^(-1)`]A. Number of mole of I(g) at equilibrium state = `5 xx 10^(-3)`B. The value of `K_(p)` at `900^(@)C = [(6.25 xx 10^(-6))/(93.84)]` atmC. Equilibrium concentration of I(g) = `12.5 xx 10^(-4)`D. Left moles of `I_(2)(g)` at equilibrium state = 0.9975 |
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Answer» Correct Answer - A::C::D `" "I_(2)(g)iff2l(g)` `"Initial 1 0"` moles `"At equilibrium 1-x 2-x"` `K_(c )=((2x)^(2))/((1-x)^(1))((1)/(V))^(Deltan_(g))` `1-x~=1" "becauseK_(c ) lt 10^(-3)` `K_(c )=(4x^(2))/(1)((1)/(4))^(1)` `x^(2)=K_(c )impliesx=sqrt(K_(c ))=25xx10^(-4)` `[1]=(2x)//4=12.5xx10^(-4)"Molar"` |
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