Reaction: A + B → C + D follow's following rate law : rate = k = \left(\left[\right. A \left]\right.\right)^{+ \frac{1}{2}} \left(\left[\right. B \left]\right.\right)^{\frac{1}{2}}. Starting with initial conc. of one mol of A and B each, what is the time taken for amount of A of become 0.25 mol. Given k = 2.31 × 10–3 sec

Engineering

Chemistry

Integrated Rate Law

Question

Reaction:

A + B → C + D

follow's following rate law :

rate = $k = {[A]}^{+ \frac{1}{2}} {[B]}^{\frac{1}{2}}$ .

Starting with initial conc. of one mol of A and B each, what is the time taken for amount of A of become 0.25 mol. Given k = 2.31 × 10^–3 sec^–1.

300 sec.

600 sec.

None of these

900 sec.

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A + B → C   + D
t = 0      1    1
‘t’ 1-x   1-x x x
rate = k[A]^½ [B]^½
– $\frac{d [A]}{dt}$ = k(1–x)^½ (1–x)^½ ⇒ – $\frac{d (1 - x)}{dt}$ = k(1 – x)   ⇒ $\frac{dx}{dt}$ = k(1 – x)
$\Rightarrow \int_{0}^{x} \frac{dx}{1 - x} = k \int_{0}^{t} dt$ ⇒ [–ln (1–x)]₀^x = kt   ⇒ –ln(1–x) = kt
⇒ –ln(1–x) = kt     ⇒ t = $\frac{1}{k} \ln (\frac{1}{1 - x})$
When [A] = 0.25
$∴$ x = 0.75
$∴ \ln \frac{1}{0 .25} = \frac{10^{3}}{2 .31} \ln 4 = \frac{2 \times 10^{3} \times 0 .693}{2 .31} = \frac{2 \times 10^{3} \times 0 .693}{2 .31} = 600 \sec .$