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))`

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 system A(g)+2B(g)hArr C(g) , the equilibrium concentration is A="0.06 mol L"^(-1), B="0.12 mol L"^(-1), C="0.216 mol L"^(-1) . The equilibrium cosntant K_(C) for the reaction is

A(g) + 3B(g) rarr 4C(g) Initially concentration of A is equal to that of B. The equilibrium concentrations of A and C are equal. Kc is :

For the system A(g)+2B(g) hArr C(g) the equilibrium concentration is A=0.06 mol L^(-1), B=0.12 mol L^(-1) C=0.216 mol L^(-1) The K_(eq) for the reaction is

The reaction A(g) + B(g)hArr2C (g) is occurred by mixing of 3 moles of A and 1 mole of B in one litre container. If a of B is (1)/(3) , then K_(c) for this reaction is:

For the equilibrium AB(g) hArr A(g)+B(g) . K_(p) is equal to four times the total pressure. Calculate the number moles of A formed if one mol of AB is taken initially.

In the reaction PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) , the equilibrium concentrations of PCl_(5) and PCl_(3) are 0.4 and 0.2 mole // litre respectively. If the value of K_(c) is 0.5 , what is the concentration of Cl_(2) in moles // litre ?