Question

An organic compound undergoes first order decomposition. The time taken for its decomposition to `1/8` and `1/10` of its initial concentration are `"t"_(1//8)` and `"t"_(1//10)` respectively. What is the value of `(["t"_(1//8)])/(["t"_(1//10)]) xx 10`? (log₁₀ 2 = 0.3)

Answer 1

For a first order process K_t = `ln  (["A"_0])/(["A"])`

where [A₀] = initial concentration

[A] = concentration of reactant remaining at time t

⇒ `"K"_("t"_(1//8)) = ln  (["A"_0])/(["A"_0]//8)`

= ln 8  

= 2.303 (log₁₀ 2) × 3  .....(∵ ln x = 2.303 log₁₀ x) .......(i)

and `"K"_("t"_(1//10)) = ln  (["A"_0])/(["A"_0]//10)`

= ln 10

= 2.303 (log₁₀ 10)  ......(ii)

As we know log 10 = 1

Now on comparing both equations (i) and (ii), we get

`("K"_("t"_(1//8)))/("K"_("t"_(1//10))) = (2.303 xx 0.3 xx 3)/(2.303)`

Thus, `("t"_(1//8))/("t"_(1//10))` = 0.3 × 3

∴ `("t"_(1//8))/("t"_(1//10)) xx 10`

= 0.3 × 3 × 10

= 9