For the reaction
`2A(g)+2B_2(g) hArr 2AB_2(g)`
the equilibrium constant `K_p` at `27^@` C is 4.0. What is the value of `K_p` for
(a)`2AB_2(g) hArr 2A(g) + 2B_2(g)`
(b) `A(g)+B_2hArr AB_2(g)`

Answer

(a)0.25 (b)2

Explanation

For the reaction , `2A(g)+2B_2(g)hArr 2AB_2(g)`
`K_p=(p_(AB_2))^2/((p_A)^2.p(B_2)^2)` =4.0
(a)`2AB_2(g) hArr 2A(g)+2B_2(g)`
`K._p=((p_A)^2(p_(B_2)))^2/(p_(AB_2))^2=1/K_p=1/4.0`
=0.25
(b)`A(g)+B_2(g) hArr AB_2(g)`
`K.._p=(p_(AB_2))^2/(p_A.pB_2)=sqrtK_p=sqrt4.0`
=2.0