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Ethylene is p...

Question

Ethylene is produced by, $C_{4} H_{8} Δ - \to 2 C_{2} H_{4}$ . The constant is $2.48 \times 10^{- 4} s e c^{- 1}$ . Time in hours (nearest integer) in which the molar ratio of the ethylene to cyclobutane in reaction mixture will attain the value of 100 is___________.

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Solution

The reaction is
$C_{4} H_{8} \to 2 C_{2} H_{4}$
Mole at $t = 0 a 0$
Mole at $t = t (a - X) 2 X$
If $\frac{2 X}{(a - X)} = 100$ i.e., $X = \frac{100 a}{102}$
$∴ t = \frac{2.303}{2.48 \times 10^{- 4}} l o g \frac{a}{(a - \frac{100 a}{102})}$
$= 15856.9 s e c o n d = 264.3 m i n \approx 4 h r$

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C_{4} H_{8}

Δ - \to

2 C_{2} H_{4}

2.303 \times 10^{- 4} s e c^{- 1} .

(l o g 5.1 = 0.7)

C_{4} H_{8}

Δ - \to

2 C_{2} H_{4}

2.303 \times 10^{- 4} s e c^{- 1} .

(l o g 5.1 = 0.7)

C_{4} H_{8} \to 2 C_{2} H_{4}

= 2.303 \times 10^{4} {sec}^{- 1}

\frac{C_{2} H_{4}}{C_{4} H_{8}}

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C_{2} H_{4} (g) + H_{2} O (l) ⟶ C_{2} H_{5} O H (l)

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