For the reaction,
`H_2(g)+CO_2(g)to CO(g)+H_2O(g)`
occurring in a closed vessel at a particular temperature. the initial concentrations of the reactants are `[H_2]=[CO_2]=1 mol*L^(-1)`. Before reaching the equilibrium, if x `mol*L^(-1)` of `H_2` has reacted then prove that : `x=(sqrt(K_c))/(1+sqrt(K_C))` where `K_c`=equilibrium constant.

For the reaction, A+B hArr C,K_c=2 at a particular temperature. Initial concentrations of A, B and C are 2mol*L^(-1),1mol*L^(-1) & 1mol*L^(-1) respectively. Express graphically the change in concentrations of A, B and C with time.

For the reaction 2H_(2)(g)+O_(2)(g)hArr 2H_(2)O(g)-K_(p)=K_(c)(RT)^(x) . Find the value of x.

Equilibrium constant, K_(c) for the reaction, N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g) at 500K is 0.061. at a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol*L^(-1)" "N_(2),2.0mol*L^(-1)" "H_(2) and 0.5mol*L^(-1)NH_(3) . is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium ?

The reaction A(g)+2BhArr2C(g)+D(g) was studied using an initial concentration of B which was 1.5 times that of A. The equilibrium concentration of A and C were found to be equal. So, K_c for the equilibrium is-

The reaction A(g)+4B(g)hArr2C(g)+3D(g) is carried out in a closed vessel of volume 2L by taking 3 mol of A(g) and 4 mol of B(g ). At equilibrium, if the amount of C(g) be 1 mol. then K_c for the reaction is-

The equilibrium constant for the reaction , 1/2H_2(g)+1/2I_2(g)toHI(g) is K_c then find equilibrium constant for 2HI(g)toH_2(g) + I_2(g)

For the reaction. H_2+I_2hArr2HI,K=47.6 . If the initial number of moles of each reactant product is 1 mol then at equilibrium-