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Electrode Potential

PCl5g → PCl3g...

Question

$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$

In the above first order reaction the concentration of $P C l_{5}$ reduces from initial concentration $50 m o l L^{- 1} t o 10 m o l L^{- 1} i n 120 minutes at 300 K$ . The rate constant for the reaction at $300 K$ is $x \times 10^{- 2} m i n^{- 1}$ . The value of $x$ is _______________.

$G i v e n : l o g 5 = 0.6989$

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Solution

$a_{0} = 50 m o l L^{- 1}$

$a_{t} = 10 m o l L^{- 1}$

For first order reaction,
$k = \frac{1}{t} l n \frac{a_{0}}{a_{t}}$

$k = \frac{1}{120} \times 2.303 l o g \frac{50}{10}$

$0.01341$

$= 1.34 \times 10^{- 2} m i n^{- 1}$

$x = 1.34 = 1 (nearest integer)$

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Similar questions

P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)

P C l_{5}

50 m o l L^{- 1} t o 10 m o l L^{- 1} i n 120 minutes at 300 K

300 K

x \times 10^{- 2} m i n^{- 1}

x

G i v e n : l o g 5 = 0.6989

P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)

P C l_{5}

50 m o l L^{- 1} t o 10 m o l L^{- 1} i n 120 minutes at 300 K

300 K

x \times 10^{- 2} m i n^{- 1}

x

G i v e n : l o g 5 = 0.6989

N_{2} O_{5} (g) \to 2 N O_{2} (g) + \frac{1}{2} O_{2} (g)

N_{2} O_{5}

\times 10^{- 2} m o l L^{- 1}

N_{2} O_{5}

1.60 \times 10^{- 2} m o l L^{- 1}

x \times 10^{- 3} m i n^{- 1}

[G i v e n : l o g 3 = 0.447, l o g 5 = 0.699]

A ⟶ B

B

0.2 {mol L}^{- 1}

x \times 10^{- 1} {mol L}^{- 1} h^{- 1}

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