Answer
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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.
Complete answer:
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).
Additional information:
Activation energy can be calculated using various methods. It can be calculated using the Arrhenius equation and also when then two temperatures and the rate constant at both temperatures are known. The temperature should be converted to kelvin while calculating activation energy using the Arrhenius equation.
Note:
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.
Complete answer:
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).
Additional information:
Activation energy can be calculated using various methods. It can be calculated using the Arrhenius equation and also when then two temperatures and the rate constant at both temperatures are known. The temperature should be converted to kelvin while calculating activation energy using the Arrhenius equation.
Note:
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.
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