For a first order reaction,the rate constant | vedclass

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(A) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{K}$.Given $t_{1/2} = 6.93 	imes 10^{-3} \, 	ext{min}$,we have $6.93

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$600$

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(A) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{K}$.Given $t_{1/2} = 6.93 	imes 10^{-3} \, 	ext{min}$,we have $6.93 	imes 10^{-3} = \frac{0.693}{K}$.Thus,$K = \frac{0.693}{6.93 	imes 10^{-3}} = 10^2 \, 	ext{min}^{-1}$.Substituting $K$ into the given equation: $\log_{10} (10^2) = 12 - \frac{6 	imes 10^3}{T}$.$2 \log_{10} 10 = 12 - \frac{6 	imes 10^3}{T}$.Since $\log_{10} 10 = 1$,we get $2 = 12 - \frac{6 	imes 10^3}{T}$.$\frac{6 	imes 10^3}{T} = 12 - 2 = 10$.$T = \frac{6 	imes 10^3}{10} = 600 \, K$.

For a first order reaction,the rate constant is given as $\log_{10} K = 12 - \frac{6 \times 10^3}{T}$. What will be the value of temperature if its half-life period is $6.93 \times 10^{-3} \, \text{min}$ (in $, K$)?

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