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

For the reaction, A_(g) + 2B_(g) iff 2C_(g) , the rate constants for the forward and the reverse reactions are 1 xx 10^-4 and 2.5 xx 10^-2 respectively. The value of equilibrium constant, K for the reaction would be

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

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

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

For the reversible reaction, A+BhArrC , the specific reaction rates for forward and reverse reactions are 1.25xx10^(3) "and" 2.75xx10^(4) respectively. The equilibrium constant for the reaction is:

For the reversible reaction, A+BhArrC , the specific reaction rates for forward and reverse reactions are 1.25xx10^(3) "and" 2.75xx10^(4) respectively. The equilibrium constant for the reaction is:

In a reversible reaction the rate constants of the forward and the backward reactions are 4.8xx10^(-5)s^(-1)and1.2xx10^(-4)s^(-1) respectively. Calculate the equilibrium constant.

Consider the reaction A(g)+B(g) hArr C(g)+D(g) Which occurs in one step. The specific rate constant are 0.25 and 5000 for the forward and reverse reaction, respectively. The equilibrium constant is

Consider the reaction A(g)+B(g) hArr C(g)+D(g) Which occurs in one step. The specific rate constant are 0.25 and 5000 for the forward and reverse reaction, respectively. The equilibrium constant is