The $t_{0.5}$ for the first-order reaction, $P C I_{5} (g) \to P C I_{2} (g) + C l_{2} (g)$ is $20$ min. The time in which the conc. of $P C I_{5}$ reduces to $25 %$ of the initial conc. is close to:

22

min

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40

min

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90

min

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50

min

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Solution

The correct option is A $40$ min
$k = \frac{2.303}{t} l o g \frac{100}{100 - x}$

$k = \frac{0.693}{t_{1 / 2}}$

$k = \frac{0.693}{20}$ = $0.03465 {m i n}^{- 1}$

$t = \frac{2.303}{0.03465} l o g \frac{100}{25}$

$t = 40 m i n$

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The t0.5 for the first-order reaction,PCI5(g)→PCI2(g)+Cl2(g) is 20 min. The time in which the conc. of PCI5 reduces to 25% of the initial conc. is close to:

The correct option is A 40 mink=2.303tlog100100−xk=0.693t1/2k=0.69320 = 0.03465 min−1t=2.3030.03465log10025t=40 min

The $t_{0.5}$ for the first-order reaction, $P C I_{5} (g) \to P C I_{2} (g) + C l_{2} (g)$ is $20$ min. The time in which the conc. of $P C I_{5}$ reduces to $25 %$ of the initial conc. is close to:

The correct option is A $40$ min
$k = \frac{2.303}{t} l o g \frac{100}{100 - x}$

$k = \frac{0.693}{t_{1 / 2}}$

$k = \frac{0.693}{20}$ = $0.03465 {m i n}^{- 1}$

$t = \frac{2.303}{0.03465} l o g \frac{100}{25}$

$t = 40 m i n$