
For a certain first-order reaction $32\% $ of the reactant is left after \[{\mathbf{570s}}\]. The rate constant of this reaction is _______ \[ \times {\text{ }}{\mathbf{1}}{{\mathbf{0}}^{-{\mathbf{3}}}}\;{{\mathbf{s}}^{\;-{\mathbf{1}}}}.\](Round off to the Nearest Integer).
[Given: \[{\mathbf{lo}}{{\mathbf{g}}_{{\mathbf{10}}}}{\mathbf{2}}{\text{ }} = {\text{ }}{\mathbf{0}}.{\mathbf{301}},{\text{ }}{\mathbf{ln10}}{\text{ }} = {\text{ }}{\mathbf{2}}.{\mathbf{303}}\]]
Answer
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Hint: Rate of a reaction is defined as the change in concentration of reactant or product with time. There are two types of rate. One is the experimental rate and the other is the theoretical rate. Theoretical rate says that \[R = - \frac{{dA}}{{dt}}\]. According to theoretical rate law, rate is directly proportional to the concentration of products raised to their respective order of the reaction. \[R = K{[A]^a}{[B]^b}\] where a and b are orders of A and B which are found experimentally and k is known as the rate constant. One equating the two types of rates and then integrating for first order reaction we come to the equation, \[kt = \ln \frac{{{a_o}}}{a}\]. A reaction in which the rate of reaction depends on the first power of the concentration of reactants is called first order reaction.
Complete Step by Step Answer:
Let ${a_o}$ be the initial concentration of reactant. Therefore, the amount of reactant left after time \[570s\] is $\frac{{32}}{{100}}$${a_o}$.
Thus, \[a = \frac{{32}}{{100}}{a_o}\]
Using formula for first order reaction, \[kt = \ln \frac{{{a_o}}}{a}\]
\[kt = \ln \frac{{{a_o}}}{{0.32{a_o}}}\]
\[kt = \ln \frac{1}{{0.32}}\]
\[k = \frac{1}{{570}}\ln \frac{1}{{0.32}}\]
$k = 0.001999$
$k = 2{(10)^{ - 3}}$s-1
Thus the correct value of integer is \[2\].
Note: In actual terms, anything that changes with time is defined as rate, be it concentration, moles, number of counts, etc. The order if not given in the question can be identified by the unit of rate constant. Unit of rate constant for a zero order reaction is \[mol{\text{ }}{L^{-1}}\;{s^{-1}}\]. It is \[{s^{-1}}\] and \[mo{l^{ - 1}}{\text{ }}L\;{s^{-1}}\] for first and second order reaction respectively. For a first order reaction, it is not necessary to know the initial concentration of the reactant as the ratio \[\frac{{{a_o}}}{a}\] cancels the term.
Complete Step by Step Answer:
Let ${a_o}$ be the initial concentration of reactant. Therefore, the amount of reactant left after time \[570s\] is $\frac{{32}}{{100}}$${a_o}$.
Thus, \[a = \frac{{32}}{{100}}{a_o}\]
Using formula for first order reaction, \[kt = \ln \frac{{{a_o}}}{a}\]
\[kt = \ln \frac{{{a_o}}}{{0.32{a_o}}}\]
\[kt = \ln \frac{1}{{0.32}}\]
\[k = \frac{1}{{570}}\ln \frac{1}{{0.32}}\]
$k = 0.001999$
$k = 2{(10)^{ - 3}}$s-1
Thus the correct value of integer is \[2\].
Note: In actual terms, anything that changes with time is defined as rate, be it concentration, moles, number of counts, etc. The order if not given in the question can be identified by the unit of rate constant. Unit of rate constant for a zero order reaction is \[mol{\text{ }}{L^{-1}}\;{s^{-1}}\]. It is \[{s^{-1}}\] and \[mo{l^{ - 1}}{\text{ }}L\;{s^{-1}}\] for first and second order reaction respectively. For a first order reaction, it is not necessary to know the initial concentration of the reactant as the ratio \[\frac{{{a_o}}}{a}\] cancels the term.
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