If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately

40 min

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50 min

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45 min

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60 min

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Solution

The correct option is C 45 min
$K = \frac{2303}{t} l o g_{10} (\frac{a}{a - x}) K = \frac{2303}{60} l o g (\frac{1}{0.4}) = \frac{2303}{60} \times 0.39 t_{\frac{1}{2}} = \frac{0.693}{k} = \frac{0.693 \times 60}{2.303 \times 0.39} = 46.15 ≃ 45 m i n$

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If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately : (log 4 = 0.60, log 5 = 0.69)

Q. If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately: (log 4 = 0.60, log 5 = 0.69)

Q. If

60 %

of a first order reaction was completed in 60 min,

50 %

of the same reaction would be completed in approximately [log 4 = 0.60, log 5 = 0.69

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If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately

The correct option is C 45 minK=2303tlog10(aa−x)K=230360log(10.4)=230360×0.39t12=0.693k=0.693×602.303×0.39=46.15≃45 min

The correct option is C 45 min
$K = \frac{2303}{t} l o g_{10} (\frac{a}{a - x}) K = \frac{2303}{60} l o g (\frac{1}{0.4}) = \frac{2303}{60} \times 0.39 t_{\frac{1}{2}} = \frac{0.693}{k} = \frac{0.693 \times 60}{2.303 \times 0.39} = 46.15 ≃ 45 m i n$