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# The first order rate constant for the decomposition of ethyl iodide by the reaction${{C}_{2}}{{H}_{5}}I(g)\to {{C}_{2}}{{H}_{4}}(g)+HI(g)$At 600 K is $(1.60)({{10}^{-5}}){{S}^{-1}}$. Activation energy is 209 KJ/mol. The rate constant of the reaction at 700K is(A).$(3.4)({{10}^{4}}){{S}^{-1}}$(B).$(4)({{10}^{-1}}){{S}^{-1}}$(C).$(6)({{10}^{3}}){{S}^{-1}}$(D).$(6.36)({{10}^{-3}}){{S}^{-1}}$  Verified
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Hint: Rate is defined as the speed at which a chemical reaction occurs. Rate is generally expressed in the terms of concentration of reactant which is consumed during the reaction in a unit of time or the concentration of product which is produced during the reaction in a unit of time.

Given in the question:
The reaction involved in the first order decomposition of ethyl iodide:
${{C}_{2}}{{H}_{5}}I(g)\to {{C}_{2}}{{H}_{4}}(g)+HI(g)$
The first order rate constant for the reaction is = $(1.60)({{10}^{-5}}){{S}^{-1}}$
Temperature given ${{T}_{1}}$= 600 K
${{T}_{2}}$= 700 K
The activation energy given: 209 KJ/mol
$\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}(\dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}})$
$\log \dfrac{{{k}_{2}}}{(1.6)({{10}^{-5}})}=\dfrac{209000}{(2.303)(8.314)}(\dfrac{1}{600}-\dfrac{1}{700})$
The value of ${{k}_{2}}=(6.39)({{10}^{-3}}){{S}^{-1}}$
The rate constant of the reaction at 700K is ${{k}_{2}}=(6.39)({{10}^{-3}}){{S}^{-1}}$

Hence the correct answer is option (D).

If the reaction is a third order reaction, the unit for third order reaction is ${{M}^{-2}}h{{r}^{-1}}or\text{ mo}{{\text{l}}^{-2}}{{L}^{2}}h{{r}^{-1}}$. The negative and positive sign in the expression of the rate or reaction only means the change in concentration. A negative charge indicates that the concentration of the reactant is decreasing, similarly a positive charge means that the concentration of product is increasing.