Equilibrium constant can also be expressed in terms of `K_(x)` , when concentrations of the species are taken in mole fraction `F_(2)(g)hArr2F(g),K_(x) = (X_(F)^(2))/(X_(F_(2)))`
For the above equilibrium mixture, aberage molar mass at 1000 K was `36.74` g `mol^(-1)` . Thus, `K_(x)` is

Write the expression for the equilibrium constant, K_C for each of the following reactions: I_2(s) +5F_2(g) hArr 2IF_5(l)

The equilibrium constant K_(c) for the following reaction at 742^(@) C is 6.90xx10^(-3) . What is K_(p) at same temperature ? (1)/(2)F_(2(g)) hArr F_(g)

The equilibrium constant K_(c) for the following reaction at 842^(@) C is 7.90xx10^(-3) . What is K_(p) at same temperature ? (1)/(2)F_(2(g)) hArr F_(g)

If F(x) = k where 'k' is constant then Delta^2f(x)=?

The equilibrium constant K_(p) for the following rection at 191^(@)C is 1.24. what is K_(c) ? B(s)+(3)/(2)F_(2)(g)hArr(g)

The equilibrium constant K_(p) for the following rection at 191^(@)C is 1.24. what is K_(c) ? B(s)+(3)/(2)F_(2)(g)hArr(g)

Passage: When all the coefficients in a balanced chemical equation are multiplied by a constant factor X the equilibrium constant (originally K) becomes K^J . Similarly, when balanced equations are added together, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for each step. Equilibrium constant of the reversed reaction is numerically equal to the reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) Consider the two reactions: XeF_(6(g))+H_(2)O_((g)) harr XeOF_(4(g))+2HF_((g)), K_(1)" "XeO_(4(g))+XeF_(6(g)) harr XeOF_(4(g))+XeO_(3)F_(2(g)), K_(2) Then the equilibrium constant for the following reaction will be XeO_(4(g))+2HF_((g)) harr XeO_(3)F_(2(g))+H_(2)O_((g))