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  • If we start a reaction from 100 mol, having the rate constant is <img alt="\mathrm{0.0693 \mathrm{~min}^{-1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B0.0693%20%5Cmath

If we start a reaction from 100 mol, having the rate constant is \mathrm{0.0693 \mathrm{~min}^{-1}}. What will be the rate of the reaction offer 20 min.

 

Option: 1

693 \mathrm{\; mol\; min^{-1}}


Option: 2

693 \mathrm{\; mol\; min^{-1}}


Option: 3

0.693 \text { mol min }^{-1}


Option: 4

\mathrm{2.5 \: mol\: min^{-1}}


Answers (1)

best_answer

\mathrm{t_{1 / x}=\frac{0.693}{k}=\frac{0.693}{0.0693}=10 \mathrm{~min}}

Reactant after 20 min = 25 mol

\mathrm{\text { Rate }\left(\frac{d A}{d t}\right)=K \mid A 3=0.0693 \times 25 \text { mol\: min}^{-1}}

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manish

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