Question

26. For a gaseous homogeneous reaction, the value
of equilibrium constant Kc is 6.15 * 10-2 mol-'L at
27°C. The value of equilibrium constant Kp
(in atm-') for the same equilibrium reaction at
same temperature is
(1) 7.5 10-1 (2) 6.15 x 10-2
(3) 2.5 x 10-3 4) 1.51
​

Answer 1

Answer:

(2)

Explanation:

For a homogeneous reaction, ∆n= 0.

So Kp= Kc= 6.15*(10)^-2

Answer 2

Value of equilibrium Kp(in atm^-1)same equilibrium reaction for the same temperature is 6.15×10^-2.

●We know that for homogeneous gaseous reaction we have

Kp=Kc(RT)^n

Where n =gaseous mole of product-gaseous mole of reactant.

●For homogeneous gaseous

H2(g)+I2(g)---->2HI(g)

So n =2-2=0

So Kp=Kc

Hence Kp=6.15×10^-2.