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Hint: To solve this question we must have a clear idea of chemical kinetics. We should also try to recall the correlation between the number of moles remaining and the passage of time in case of a first order reaction.
Complete step by step solution:
To determine the order of a reaction and the value of the rate constant from experiments we generally use integrated rate laws.
For the general reaction,
\[aA\, + bB \to C\] having no intermediate steps in its mechanism, the rate law is given by:
\[r = k{[A]^x}{[B]^y}\]
In this equation [A] and [B] expresses the concentrations of the reactants, in units of moles per liter. The exponents x and y depend solely on the reaction, and can only be determined experimentally; they are not related to the stoichiometric coefficients in the chemical equation. Furthermore, k is the rate constant of the reaction.
We should remember that in a first-order reaction the rate depends linearly on only one reactant concentration.
Hence, the formula to determine the rate constant of a first order reaction is:
\[k = \frac{{2.303}}{t}\log \frac{a}{{a - x}}\]
Where ‘t’ is the time elapsed, ‘a’ is the initial concentration and ‘x’ is the concentration after time t.
Substitution of the values in this equation should be by considering the initial concentration to be 100.
Therefore, for 20% completion, the remaining reactant after time t is 100 – 20 = 80.
\[k = \frac{{2.303}}{{10}}\log \frac{{100}}{{100 - 20}} = 0.0223\,{\min ^{ - 1}}\]
Hence, the correct answer is Option (A) $0.0223\,{\min ^{ - 1}}$
Additional Information:
If we plot the natural logarithm of the concentration of a reactant (ln) versus time, a reaction that has a first-order rate law will produce a straight line, whereas a reaction of any other order will not yield a straight line.
Note: We must remember that for first-order reactions, the half-life is independent of the initial concentration of the reactant, which is a unique feature of first-order reactions. This means that it takes the same time for a reactant to decrease from 1 M to 0.5 M concentration as it takes for it to decrease from 0.1 M to 0.05 M.
Complete step by step solution:
To determine the order of a reaction and the value of the rate constant from experiments we generally use integrated rate laws.
For the general reaction,
\[aA\, + bB \to C\] having no intermediate steps in its mechanism, the rate law is given by:
\[r = k{[A]^x}{[B]^y}\]
In this equation [A] and [B] expresses the concentrations of the reactants, in units of moles per liter. The exponents x and y depend solely on the reaction, and can only be determined experimentally; they are not related to the stoichiometric coefficients in the chemical equation. Furthermore, k is the rate constant of the reaction.
We should remember that in a first-order reaction the rate depends linearly on only one reactant concentration.
Hence, the formula to determine the rate constant of a first order reaction is:
\[k = \frac{{2.303}}{t}\log \frac{a}{{a - x}}\]
Where ‘t’ is the time elapsed, ‘a’ is the initial concentration and ‘x’ is the concentration after time t.
Substitution of the values in this equation should be by considering the initial concentration to be 100.
Therefore, for 20% completion, the remaining reactant after time t is 100 – 20 = 80.
\[k = \frac{{2.303}}{{10}}\log \frac{{100}}{{100 - 20}} = 0.0223\,{\min ^{ - 1}}\]
Hence, the correct answer is Option (A) $0.0223\,{\min ^{ - 1}}$
Additional Information:
If we plot the natural logarithm of the concentration of a reactant (ln) versus time, a reaction that has a first-order rate law will produce a straight line, whereas a reaction of any other order will not yield a straight line.
Note: We must remember that for first-order reactions, the half-life is independent of the initial concentration of the reactant, which is a unique feature of first-order reactions. This means that it takes the same time for a reactant to decrease from 1 M to 0.5 M concentration as it takes for it to decrease from 0.1 M to 0.05 M.
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