The equilibrium constant for the reactions
`N_(2)+O_(2)hArr2NO"and "NO+1/2O_(2)hArrNO_(2)"are"K_(1)"and"K_(2)` respectively, that will be the equilibrium constant for the reaction

Equilibrium constants for (a) N_2 + O_2 hArr 2NO and (b) NO + 1/2O_2 hArr NO_2 are K_1 and K_2 , respectively, then the equilibrium constant K_3 for the reaction N_2 + 2O_2 hArr 2NO_2 will be :-

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 equilibrium constant for the reaction, Ag_2O(s) hArr2Ag(s) +1/2 O_2 (g) is given by

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

The equilibrium constant for the given reaction is 100. N_(2)(g)+2O_(2)(g) hArr 2NO_(2)(g) What is the equilibrium constant for the reaction ? NO_(2)(g) hArr 1//2 N_(2)(g) +O_(2)(g)

The value of the equilibrium constant for the reaction : H_(2) (g) +I_(2) (g) hArr 2HI (g) at 720 K is 48. What is the value of the equilibrium constant for the reaction : 1//2 H_(2)(g) + 1//2I_(2)(g) hArr HI (g)

If the equilibrium constant for the given reaction is 0.25 NOhArr(1)/(2)N_(2)+(1)/(2)O_(2) , then the equilibrium constant for the reaction (1)/(2)N_(2)+(1)/(2)O_(2)hArrNO will be

When the reaction, 2NO_(2)(g) hArr N_(2)O_(4)(g) reaches equilibrium at 298 K . The partial pressure of NO_(2) and N_(2)O_(4) are 0.2 Kpa and 0.4 Kpa , respectively. What is the equilibrium constant K_(p) of the above reaction at 298 K ?

The equilibrium constant for the reaction N_(2)+2O_(2)hArr2NO_(2) at a particular temperature is 100 . Write down the equilibrium law equations for the following reaction and determine the values of equilibrium constants. 2NO_(2)hArrN_(2)+2O_(2) NO_(2)hArr1//2N_(2)+O_(2)

Write the equilibrium constant expressions for the following reactions. N_2(g) +O_2(g)hArr 2NO(g)