90% of first order process X →Y is completed in a time equals to 99% of another first order process Y →Q . If K value of Y →Q is 0.09sec1,K valueof X→Ywill be (in sec-1)

K=2.303tlog [R0R]K=2.303tlog [1001]K=2.303tlog [100]K=2.303tlog [102]K=2.303×2t0.09=2.303×2t t =4.6060.09=511.77K=2.303×10511.770.0045 sec-1

90% of first order process X →Y is completed in a time equals to 99% of another first order process Y →Q . If K value of Y →Q is 0.09sec1,K valueof X→Ywill be (in sec-1)

K=2.303tlog [R0R]K=2.303tlog [1001]K=2.303tlog [100]K=2.303tlog [102]K=2.303×2t0.09=2.303×2t t =4.6060.09=511.77K=2.303×10511.770.0045 sec-1

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90% of first order process X $\to$ Y is completed in a time equals to 99% of another first order process Y $\to$ Q . If K value of Y $\to$ Q is ${0.09sec}^{1}, K valueof X \to Y$ will be (in sec^-1)

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0.27

0.3

0.03

0.045

answer is D.

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Detailed Solution

$K = \frac{2.303}{t} \log [\frac{R_{0}}{R}]$

$K = \frac{2.303}{t} \log [\frac{100}{1}]$

$K = \frac{2.303}{t} \log [100]$

$K = \frac{2.303}{t} \log [10^{2}]$

$K = \frac{2.303 \times 2}{t}$

$0.09 = \frac{2.303 \times 2}{t} t = \frac{4.606}{0.09} = 511.77$

$K = \frac{2.303 \times 10}{511.77}$

$0.0045 \sec^{- 1}$

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90% of first order process X $\to$ Y is completed in a time equals to 99% of another first order process Y $\to$ Q . If K value of Y $\to$ Q is ${0.09sec}^{1}, K valueof X \to Y$ will be (in sec^-1)