`K_(p)` is equal to `K_(c)` if `Deltan` is positive.

For the following equilibrium N_(2)O_(4)(g)hArr 2NO_(2)(g) K_(p) is found to be equal to K_(c) . This is attained when :

Unit of K_(p) is (atm)^(Deltan)

If K_(c) is not numerically equal to K_(p) , how can both of the following equations be valid? DeltaG^(ɵ)=-2.303 RT log K_(c), DeltaG^(ɵ)=-2.303 RT log K_(p)

The relation between K_(p) and K_(c) is K_(p)=K_(c)(RT)^(Deltan) unit of K_(p)=(atm)^(Deltan) , unit of K_(c)=(mol L^(-1))^(Deltan) Given: 2NO(g)+O_(2)(g) hArr 2NO_(2)(g), K_(1) 2NO_(2)(g) hArr N_(2)O_(4)(g), K_(2) 2NO(g)+O_(2)(g) hArr N_(2)O_(4)(g), K_(3) Which of the following relations is correct?

The relation between K_(p) and K_(c) is K_(p)=K_(c)(RT)^(Deltan) unit of K_(p)=(atm)^(Deltan) , unit of K_(c)=(mol L^(-1))^(Deltan) Consider the following reactions: i. CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g), K_(1) ii. CH_(4)(g)+H_(2)O(g) hArr CO(g)+3H_(2)(g), K_(2) iii. CH_(4)(g)+2H_(2)O(g) hArr CO_(2)(g)+4H_(2)(g), K_(3) Which of the following is correct?

The relation between K_(p) and K_(c) is K_(p)=K_(c)(RT)^(Deltan) unit of K_(p)=(atm)^(Deltan) , unit of K_(c)=(mol L^(-1))^(Deltan) H_(3)ClO_(4) is a tribasic acid, it undergoes ionisation as H_(3)ClO_(4) hArr H^(o+)+H_(2)ClO_(4)^(-), K_(1) H_(2)ClO_(4)^(-) hArr H^(o+)+HClO_(4)^(2-), K_(2) HClO_(4)^(2-) hArr H^(o+)+ClO_(4)^(3-), K_(3) Then, equilibrium constant for the following reaction will be: H_(3)ClO_(4) hArr 3H^(o+)+ClO_(4)^(3-)

The relation between K_(p) and K_(c) is K_(p)=K_(c)(RT)^(Deltan) unit of K_(p)=(atm)^(Deltan) , unit of K_(c)=(mol L^(-1))^(Deltan) The equilibrium constant of the following reactions at 400 K are given: 2H_(2)O(g) hArr 2H_(2)(g)+O_(2)(g), K_(1)=3.0xx10^(-13) 2CO_(2)(g) hArr 2CO(g)+O_(2)(g), K_(2)=2xx10^(-12) Then, the equilibrium constant K for the reaction H_(2)(g)+CO_(2)(g) hArr CO(g)+H_(2)O(g) is

K_(p) " and " K_(c) are related to each other as :