A→Product [A]0=2M. After 10 minutes reaction is 10% completed if − dAdt=k[A] then T50% is approximately.

k=2.30310log10090=0.0105 min−1T50%=0.693k=0.6930.0105=65.7 min−1

A→Product [A]0=2M. After 10 minutes reaction is 10% completed if − dAdt=k[A] then T50% is approximately.

k=2.30310log10090=0.0105 min−1T50%=0.693k=0.6930.0105=65.7 min−1

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$A \to \Pr o d u c t {[A]}_{0} = 2 M .$ After 10 minutes reaction is 10% completed if $- \frac{d A}{d t} = k [A]$ then $T_{50 %}$ is approximately.

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$69.3 \min$

$0.0693 \min$

$66.0 \min$

$0.693 \min$

answer is C.

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$k = \frac{2.303}{10} \log \frac{100}{90} = 0.0105 \min^{- 1}$

$T_{50 %} = \frac{0.693}{k} = \frac{0.693}{0.0105} = 65.7 \min^{- 1}$

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A→Product [A]0=2M. After 10 minutes reaction is 10% completed if − dAdt=k[A] then T50% is approximately.

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$A \to \Pr o d u c t {[A]}_{0} = 2 M .$ After 10 minutes reaction is 10% completed if $- \frac{d A}{d t} = k [A]$ then $T_{50 %}$ is approximately.