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(D) For a $1^{st}$ order reaction:$k = \frac{2.303}{t} \log \frac{a_0}{a_t}$Given:$a_0 = 20 \ mmol$$a_t = 10 \ mmol$$t = 1.151 \ min$Substituting the

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$0.60$

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(D) For a $1^{st}$ order reaction:$k = \frac{2.303}{t} \log \frac{a_0}{a_t}$Given:$a_0 = 20 \ mmol$$a_t = 10 \ mmol$$t = 1.151 \ min$Substituting the values:$k = \frac{2.303}{1.151} \log \left( \frac{20}{10} ight)$$k = \frac{2.303}{1.151} 	imes \log 2$$k = \frac{2.303 	imes 0.3010}{1.151}$$k = 0.60 \ min^{-1}$Alternatively,since the concentration reduces to half ($20 \ mmol$ to $10 \ mmol$),the time taken is the half-life $(t_{1/2})$:$t_{1/2} = 1.151 \ min$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1.151} \approx 0.60 \ min^{-1}$

In a first order reaction,the concentration of the reactant decreases from $20 \ mmol$ to $10 \ mmol$ in $1.151 \ min$. What is the rate constant (in $min^{-1}$)?

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