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Question: Calculate the half-life of a first-order reaction from their rate of constants given below: (a) \(...

Calculate the half-life of a first-order reaction from their rate of constants given below:
(a) 200s1200{{s}^{-1}} ; (b) 2min12{{\min }^{-1}} ; (c) 4year14yea{{r}^{-1}}.

Explanation

Solution

The time taken for the concentration of a given reactant to reach 50% of its initial concentration is known as the half-life of a chemical reaction.

Complete step by step solution:
-In other words, the time period taken for the reactant concentration to reach half of its initial value is known as the half-life of a chemical reaction.
-The half-life of a chemical reaction is denoted by t1/2{{t}_{1/2}}, and i'd usually express it in seconds.
-For the first-order reaction, the rate constant can be mathematical as-
k=2.303tlog[R]0[R]k=\dfrac{2.303}{t}\log \dfrac{{{[R]}_{0}}}{[R]}
From the definition of half-life, substituting t=t1/2;[R]=[R]0/2t={{t}_{1/2}};[R]={{[R]}_{0}}/2 in the expression for the first-order rate constant, we will get,
k=2.303t1/2log[R]0[R]0/2k=\dfrac{2.303}{{{t}_{1/2}}}\log \dfrac{{{[R]}_{0}}}{{{[R]}_{0}}/2}
Rearranging the above expression, we will get
t1/2=2.303klog(2){{t}_{1/2}}=\dfrac{2.303}{k}\log (2)
t1/2=0.693k\Rightarrow {{t}_{1/2}}=\dfrac{0.693}{k}
-Using the above formula for the calculation of half-lives,
(a) t1/2=0.693200=3.465×103{{t}_{1/2}}=\dfrac{0.693}{200}=3.465\times {{10}^{-3}}s (appx.)
(b) t1/2=0.6932=0.3465{{t}_{1/2}}=\dfrac{0.693}{2}=0.3465mins (appx.)
(c) t1/2=0.6934=0.1732{{t}_{1/2}}=\dfrac{0.693}{4}=0.1732 years (appx.)

Note: -A chemical reaction that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters (normally rate coefficients and partial reaction orders) is known as the rate law or rate equation.
-A first-order reaction is the reaction that depends on the concentration of only one reactant. The first-order reaction is also referred to as unimolecular reaction, other reactants can be present but will be present in zero order.
-The rate law for the first-order reaction is given as-
d[A]dt=k[A]-\dfrac{d[A]}{dt}=k[A]
-In organic chemistry, all the reactions which belong to the class of nucleophilic substitution reactions (SN1)({{S}_{N}}1)consists of first-order reactions. For example, in the reaction of aryldiazonium ions with nucleophiles in aqueous solution, ArN2++XArX+N2Ar{{N}_{2}}^{+}+{{X}^{-}}\to ArX+{{N}_{2}}, the rate equation for this reaction is v=k[ArN2+]v=k[Ar{{N}_{2}}^{+}], where Ar is an aryl group.
-Some examples of first-order reactions are given below-
(i) H2O2(l)H2O(l)+12O2  \begin{aligned} & {{H}_{2}}{{O}_{2}}(l)\to {{H}_{2}}O(l)+\dfrac{1}{2}{{O}_{2}} \\\ & \\\ \end{aligned}
(ii) SO2Cl2(l)SO2(g)+Cl2(g)S{{O}_{2}}C{{l}_{2}}(l)\to S{{O}_{2}}(g)+C{{l}_{2}}(g)
(iii) 2N2O5(g)4NO2(g)+O22{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}