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Hint: At half-life period, the concentration of the reactant would be exactly half of the initial amount present. We can determine the half-life period if we put this value in the rate law expression.
Complete step by step answer:
Rate law states that $\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log }\frac{\text{a}}{\text{a-x}}$
Where, t = time taken for reaction
a = initial concentration of the reactant
a-x = final concentration
Half life of reaction is the time required for the concentration of the reactant to reach exact half of the initial amount present. So, at this time the concentration of the reactant would be ${}^{\text{a}}/{}_{\text{2}}$ and the time would be ${{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}$.
So, if we put these values in the rate law expression we get,
\[\text{k = }\frac{\text{2}\text{.303}}{{{\text{t}}_{{}^{1}/{}_{2}}}}\text{ log }\frac{\text{a}}{{}^{\text{a}}/{}_{\text{2}}}\]
\[\text{ = }\frac{\text{2}\text{.303}}{{{\text{t}}_{{}^{1}/{}_{2}}}}\text{ log 2}\]
\[\text{k = }\frac{0.693}{{{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}}\] \[\therefore \text{ }{{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}\text{ = }\frac{0.693}{\text{k}}\]
We have been given that \[\text{ k = 200 }{{\text{s}}^{-1}}\]. Substituting this value in above expression we get,
\[\text{ }{{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}\text{ = }\frac{0.693}{200}\] \[\text{ = 3}\text{.46 }\times \text{ 1}{{\text{0}}^{-3}}\text{ s}\]
Hence, option B is correct.
Additional information: The rate of reaction or reaction rate is the speed at which reactants are converted into products. Different factors such as concentration of reactant and product, pressure, temperature, solvent, presence of catalyst and order of reaction have a drastic effect on the rate of reaction.
The power dependence of rate on the concentration of all reactants is called the order of the reaction. When the rate of the reactions depends on the concentration of only one reactant the order of reaction is 1.
Note:
The formula of half life used here i.e. ${{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}\text{ = }\frac{0.693}{\text{k}}$. is applicable only to first order reactions and not reactions of second, third or zero order.
Complete step by step answer:
Rate law states that $\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log }\frac{\text{a}}{\text{a-x}}$
Where, t = time taken for reaction
a = initial concentration of the reactant
a-x = final concentration
Half life of reaction is the time required for the concentration of the reactant to reach exact half of the initial amount present. So, at this time the concentration of the reactant would be ${}^{\text{a}}/{}_{\text{2}}$ and the time would be ${{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}$.
So, if we put these values in the rate law expression we get,
\[\text{k = }\frac{\text{2}\text{.303}}{{{\text{t}}_{{}^{1}/{}_{2}}}}\text{ log }\frac{\text{a}}{{}^{\text{a}}/{}_{\text{2}}}\]
\[\text{ = }\frac{\text{2}\text{.303}}{{{\text{t}}_{{}^{1}/{}_{2}}}}\text{ log 2}\]
\[\text{k = }\frac{0.693}{{{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}}\] \[\therefore \text{ }{{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}\text{ = }\frac{0.693}{\text{k}}\]
We have been given that \[\text{ k = 200 }{{\text{s}}^{-1}}\]. Substituting this value in above expression we get,
\[\text{ }{{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}\text{ = }\frac{0.693}{200}\] \[\text{ = 3}\text{.46 }\times \text{ 1}{{\text{0}}^{-3}}\text{ s}\]
Hence, option B is correct.
Additional information: The rate of reaction or reaction rate is the speed at which reactants are converted into products. Different factors such as concentration of reactant and product, pressure, temperature, solvent, presence of catalyst and order of reaction have a drastic effect on the rate of reaction.
The power dependence of rate on the concentration of all reactants is called the order of the reaction. When the rate of the reactions depends on the concentration of only one reactant the order of reaction is 1.
Note:
The formula of half life used here i.e. ${{\text{t}}_{{}^{\text{1}}/{}_{\text{2}}}}\text{ = }\frac{0.693}{\text{k}}$. is applicable only to first order reactions and not reactions of second, third or zero order.
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