Question

For a reaction: \[\ce{A_{(g)} -> nB_{(g)}}\] the rate constant is 6.93 × 10^-4 sec^-1. The reaction is performed at constant pressure and temperature of 24.63 atm and 300 K, starting with 1 mole of pure 'A'. If concentration of 'B' after 2000 sec is `3/3.25` M then the value of 'n' is ______.

Options

• 0

• 2

• 1

• 4

Answer 1

For a reaction: \[\ce{A_{(g)} -> nB_{(g)}}\] the rate constant is 6.93 × 10^-4 sec^-1. The reaction is performed at constant pressure and temperature of 24.63 atm and 300 K, starting with 1 mole of pure 'A'. If concentration of 'B' after 2000 sec is `3/3.25` M then the value of 'n' is 4.

Explanation:

\[\ce{A_{(g)} -> nB_{(g)}}\]

t = 1        0

t = 1 - x   nx

n_T = 1 + (n - 1)x

Kt = `n "A"_0/"A"_"t"`

`=> 0.693 xx 10^-3 xx "t" = l "n"  1/"n"_"t"`

ln 2 × 2 = `l "n" 1/"n"_"t"`

`=> "n"_"t" = 1/4 = 0.25`

`"V"_1/"n"_1 = "V"_2/"n"_2 [24.63 xx "V" = 1 xx 24.63]`

`1/1 = "V"_2/(1/4 + 3/4 "n")`

(B) = `(3/4)/(1/4 + 3/4"n") = 3/3.25`

`therefore (3/4)/(1/4 + 3/4 "n") = 3/3.25`

3.25 n = 1 + 3n

n = 4