For the reaction equilibrium , `N_2O_4(g) hArr 2NO_(2)(g)` the concentrations of `N_2O_4` and`NO_2` at equilibrium are `4.8xx10^(-2)` and `1.2xx10^(-2) mol L^(-1)` respectively.The value of `K_C` for the reaction is :

For the reaction equilibrium, N_2O_4 (g) hArr 2NO_2 (g) the concentrations of N_2O_4 and NO_2 at equilibrium are 4.8 xx 10^(-2) and 1.2 xx 10 ^(-2) " mol " L^(-1) respectively . The value of K_c for the reaction is

For the reaction equilibrium, N_(2)O_(4(g))hArr2NO_(2(g)) , the concentration of N_(2)O_(4) and NO_(2) at equilibrium are 4.8xx10^(-2) and 1.2xx10^(-2) mol/L respectively. The value of K_(c) for the reaction is:

For an equilibrium reaction, N_(2)O_(4)(g) hArr 2NO_(2)(g) , the concentrations of N_(2)O_(4) and NO_(2) at equilibrium are 4.8 xx 10^(-2) and 1.2 xx 10^(-2) mol//L respectively. The value of K_(c) for the reaction is

For a reaction equilibrium, N_2O_4 ​ (g)⇌2 NO_2 ​ (g), the concentrations of N2O4 and NO2 ​ at equilibrium are 3.6 and 1.2 mol/L respectively. The value of K c ​ , for the reaction is:

Calculate the value of equillibrium constant, N2 O4 ​ (g)→2NO 2 ​ (g), the concentration of N 2 ​ O 4 ​ (g) and NO 2 ​ at equilibrium are 4.8×10 −2 and 1.2×10 −2 mol/L respectively.

For the reaction : N_(2)O_(4)(g)hArr 2NO_(2)(g) the concentration of the equilibrium mixture at 300 K are [N_(2)O_(4)]=4.8xx10^(-2)mol L^(-1) and [NO_(2)]=1.2xx10^(-2)mol L^(-1). K_(c ) for the reaction is :

For the reaction, N_(2)O_(4)(g) hArr 2NO_(2)(g) , the concentration of an equilibrium mixture at 298 K is N_(2)O_(4)=4.50xx10^(-2) mol L^(-1) and NO_(2)=1.61xx10^(-2) mol L^(-1) . What is the value of equilibrium constant?