For a reaction, `A+B^(+2)rarr B+A^(+2)`;
`E^(@)=0.2955V`. Hence equilibrium constant of the reaction is at `25^(@)C`:
`10`
`10^(10)`
`-10`
`10^(-10)`

For a reaction A (s) + B^(2+) rarr B(s) + A ^(2+) , at 25^(@) C E^(o) "" _(cell) = 0.2955 V. Hence, K_(eq) will be :

An equilibrium constant of 10^(-4) for a reaction means, the equilibrium is

For an exothermic reaction , temperature increase by 10^@C , the equilibrium constant will

The value of K_(c ) for the reaction 3O_(2)(g) hArr 2O_(3)(g) is 2.0xx10^(-50) at 25^(@)C . If the equilibrium concentration of O_(2) in air at 25^(@)C is 1.6xx10^(-2) , what is the concentration of O_(3) ?

For the reaction Br_(2) hArr 2Br , the equilibrium constants at 327^(@)C and 527^(@)C are, respectively, 6.1xx10^(-12) and 1.0xx10^(-7) . What is the nature of the reaction?

The equilibrium constant for the reaction A(g)hArrB(g) and B(g)hArrC(g) are K_1=10^(-2) and K_2=10^(-1) respectively . Equilibrium constant for the reaction (1/2)C(g)hArr(1/2)A(g) is

1 mole of a gas A is taken in a vessel of volume 1L . It dissociates according to the reaction A(g)hArrB(g)+C(g) "at" 27^(@)C . Forward and backward reaction rate constants for the reaction are 1.5xx10^(-2) "and" 3xx10^(-2) respectively. Find the concentrations of A,B "and" C at equilibrium.

The rate constant of a reaction is given by k = 2.1 xx 10^(10) exp(-2700//RT) . It means that

For a reaction A(s) + 2B^(+) rarr A^(2+)+2B(s). K_(c) has been found to be 10^(14) The E^(o) ""_(cell) is :

For the reaction, A(g)+2B(g) hArr 2C(g) , the rate constant for forward and the reverse reactions are 1xx10^(-4) and 2.5xx10^(-2) respectively. The value of equilibrium constant, K for the reaction would be