For a reaction, `A`+`B^(2+)``rarr` `B`+`A^(2+)`; `E^(circ)`=0.2955V. Hence equilibrium constant of the reaction at `25^(circ) C`: (1) 10 (2) `10^10` (3) -10 (4) `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

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

The fraction equivalent to 1 2/3 is (a) (10)/3 (b) 3/5 (c) (10)/6 (d) 6/(10)

The equilibrium constants for the reaction, A_2hArr2A A at 500K and 700K are 1xx10^(-10) and 1xx10^(-5) . The given reaction is

A certain weak acid has a dissocation constant of 1.0 xx 10^(-4) . The equilibrium constant for its reaction with a strong base is

For a reaction: A(s) + 2B ^(+) rarr A^(2+) +2 B (s). K_(aq) has been found to be 10^(12) The E^(o) cell is :

The equilibrium constants for the reaction Br_(2)hArr 2Br at 500 K and 700 K are 1xx10^(-10) and 1xx10^(-5) respectively. The reaction is:

The equilibrium constants for the reaction Br_(2)hArr 2Br at 500 K and 700 K are 1xx10^(-10) and 1xx10^(-5) respectively. The reaction is:

For a reaction A(s)+2B^(o+) rarr A^(2+)+2B K_(c) has been found to be 10^(12) . The E^(c-)._(cell) is