(i)

Sol. A B C ; consecutive reaction

(ii)

Sol. Sol. = k a

or ln = k a t

or t 1/2 =

or k a =

From the plot t 1/2 for the disappearance of A is 0.125 min

∴ k a = min –1 = 5.544 min –1

(iii)

Sol. = k a – k b

or + k b = k a = k a 0

Multiplying both sides with we get

or + k b = k a 0

or = k a 0

On integration

= + C

when t = 0, = 0 and = 0

∴ C =

or =

At maximum concentration of B, = 0

∴ = = 0

or =

or =

or ln = (k b – k a )t

or (k b – k a )t max = ln

or t max = ln

when t = 1 min, = 0.5 M

∴ 0.5 = …(I)

when t = 2 min = 0.2 (M)

0.2 = …(II)

(II) ÷ (I)

=

0.4 =

or 1.56 × 10 –3 – 0.4 + – 1.529 × 10 –5 = 0

or – 2 × × 0.2 + (0.2) 2 ~ (0.2) 2 – 1.56 × 10 –3 or = 0.04156 = (0.204)

2

or – 0.2 = 0.204

or = 0.2 + 0.204 = 0.404

or – k b = –0.906

or k b = 0.906 min –1 (iv)

Sol. t max = ln

k a = 5.544 min

–1 and k b = 0.906 min –1 ∴ t max = ln

or t max = × 1.8114 = 0.39 mins

(v)

Sol. + + = 0

or = 0 – –

= 1 – – 0

= 1 – – [since 0 = 1 M]

= 1 –

= 1 –

= 1 +