In the first order reaction the concentration of reactant decreases from $1.0 M$ to $0.25 M$ in $20$ minutes. The value of specific rate is:

69.32

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6.932

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0.6932

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0.06932

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Solution

The correct option is D $0.06932$
$\frac{2.303}{20} l o g \frac{2}{0.5} = 0.06932$ .

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In the first order reaction the concentration of reactant decreases from 1.0 M to 0.25 M in 20 minutes. The value of specific rate is:

The correct option is D 0.069322.30320log20.5=0.06932.

In the first order reaction the concentration of reactant decreases from $1.0 M$ to $0.25 M$ in $20$ minutes. The value of specific rate is:

The correct option is D $0.06932$
$\frac{2.303}{20} l o g \frac{2}{0.5} = 0.06932$ .