The value of `K_(p)` for the reaction `2H_(2)S(g) hArr 2H_(2)(g) +S_(2)(g) is 1.2 xx 10^(-2)` at `1065^(circ)C`. The value of K, for this reaction is

Options

A. `1.2 xx 10^(-2)`
B. `lt 1.2 xx 10^(-2)`
C. 82
D. `gt 1.2 xx 10^(-2)`

Answer

B

Explanation

`K_(p) = 1.2 xx 10^(-2)`
`T = 1065^(0)C = 1065+273 = 1338 K`
`:. K_(p) = K_(c)(RT)^(Delta n)`
Here, R = 0.083
`Delta n` = Number of moles of products - Number of moles of reactants = (2+1)-2 = 3-2 = 1
`:. 1.2 xx 10^(-2) = K_(c)(0.083 xx 1338)`
`:. K_(c) = 1.2 xx 10^(-2)/(0.083 xx 1338) = 1.0 xx 10^(-4) i.e., K_(c) < 1.2 xx 10^(-2)`