Question

In Freundlich adsorption isotherm, slope of AB line is:

विकल्प

• `1/"n" "with" (1/"n" = "0 to 1")`

• `log  1/"n"` with (n < 1)

• log n with (n > 1)

• n with (n, 0.1 to 0.5)

Answer 1

`bb(1/"n" "with" (1/"n" = "0 to 1"))`

Explanation:

According to Freundlich adsorption isotherm, `"x"/"m" = "KP"^(1/"n")` (where n and K are constant depending upon the nature of adsorbate and adsorbent, n is always greater than unity)

⇒ `log  "x"/"m" = log "K" + 1/"n" log "p"`

Therefore, slope is `1/"n"` with `(1/"n" = "0 to 1")`, `0 < 1/"n" < 1`.