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For $A \to B$ , $\frac{k_{t + 10}}{k_{t}} = 2.$ If the rate constant at 300 k is $‘ Y ’ {min}^{- 1},$ at what temperature rate constant becomes $‘ Y ’ {min}^{- 1}$ ?

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$57^{0} C$

$320^{0} C$

$280 K$

$\sqrt{9 \times 300} K$

answer is A.

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Detailed Solution

$μ = 2 \frac{K_{2}}{K_{1}} = μ^{x} \frac{8 y}{y} = 2^{x} \Rightarrow x = 3 x = (\frac{T_{2} - T_{1}}{10}) \Rightarrow 3 = \frac{T_{2} - 300}{10} T_{2} = 330 K \Rightarrow 57^{0} C$

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