Ethylene is produced by : $C_{4} H_{8}$ $Δ - \to$ $2 C_{2} H_{4}$ The rate constant is $2.303 \times 10^{- 4} s e c^{- 1} .$ In what time (in seconds), the molar ratio of the ethylene to cyclobutane in reaction mixture attain the value of 100. $(l o g 5.1 = 0.7)$

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Solution

$C_{4} H_{8}$ $Δ - \to$ $2 C_{2} H_{4}$
Mole at t = 0 a 0
Mole at t = t a-x 2x

When $\frac{2 x}{a - x} = 100$ , then $x = \frac{100}{102} a, t = \frac{2.303}{2.303 \times 10^{- 4}} l o g_{10} \frac{a}{a - \frac{100 a}{102}} = 10^{4} l o g 51 = 10^{4} \times (1 + 0.7) = 17000 s$

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Q. Ethylene is produced by :

C_{4} H_{8}

Δ - \to

2 C_{2} H_{4}

The rate constant is

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

In what time (in seconds), the molar ratio of the ethylene to cyclobutane in reaction mixture attain the value of 100.

(l o g 5.1 = 0.7)

Q. 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___________.

Q. Ethylene is produced by,

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

; rate constant

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

, After what time the molar ratio of

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

attain the value

1.

(a) Calculate the number of moles and number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?
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Ethylene is produced by : C4H8 Δ−→ 2C2H4The rate constant is 2.303×10−4sec−1. In what time (in seconds), the molar ratio of the ethylene to cyclobutane in reaction mixture attain the value of 100. (log5.1=0.7)

C4H8 Δ−→ 2C2H4 Mole at t = 0 a 0 Mole at t = t a-x 2x When 2xa−x=100, then x=100102a,t=2.3032.303×10−4log10aa−100a102=104log51=104×(1+0.7)=17000 s

Ethylene is produced by : $C_{4} H_{8}$ $Δ - \to$ $2 C_{2} H_{4}$ The rate constant is $2.303 \times 10^{- 4} s e c^{- 1} .$ In what time (in seconds), the molar ratio of the ethylene to cyclobutane in reaction mixture attain the value of 100. $(l o g 5.1 = 0.7)$

$C_{4} H_{8}$ $Δ - \to$ $2 C_{2} H_{4}$
Mole at t = 0 a 0
Mole at t = t a-x 2x

When $\frac{2 x}{a - x} = 100$ , then $x = \frac{100}{102} a, t = \frac{2.303}{2.303 \times 10^{- 4}} l o g_{10} \frac{a}{a - \frac{100 a}{102}} = 10^{4} l o g 51 = 10^{4} \times (1 + 0.7) = 17000 s$