For a reaction : `A(g)+2B(g)hArrC(g)+D(g)+E(g), K_(eq)=2`
Calculate equilibrium concentration of B if initially 10 moles of A are mixed with 20 moles of b in a 2 litre rigid container.

For the given reaction, [A(g)+2B(g)rarr3C(g)] , Find the equilibrium concentration of "B" (mol "/L" ) in "1/8" litre container if two moles of "A" and four moles of "B" are taken initially.Given " (K_(c)=2times10^(-3))

2 mol of C(g) and 4 mole of D(g) are taken in a closed rigid container of 1 litre volume and allowed to attain equilibrium as : A(g)+2B(g)hArr C(g)+D(g),K_C=10^(-8) Calculate the equilibrium concentration of C(g).(If your answer in scientific notation is x xx10^(-y) then fill y).

For a reaction A_((g)) + B_((g))hArrC_((g)) + D_((g)) the intial concentration of A and B are equals but the equilibrium constant of C is twice that of equilibrium concentration of A. Then K is

For the reaction 2A(g)+B(g)hArr C(g)+D(g), K_c=10^(12) .if initially 4,2,6,2 moles of A,B,C,D respectively are taken in a 1 litre vessel, then the equilibrium concentration of A is :

For the reaction 2A(g)+B(g)hArr C(g)+D(g), K_c=10^(12) .if initially 4,2,6,2 moles of A,B,C,D respectively are taken in a 1 litre vessel, then the equilibrium concentration of A is :

For the reaction 2A(g)+B(g)hArr C(g)+D(g), K_c=10^(12) .if initially 4,2,6,2 moles of A,B,C,D respectively are taken in a 1 litre vessel, then the equilibrium concentration of A is :