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Question: Rate constant k of a reaction varies with temperature according to the equation \[{\text{log}}\,{\te...

Rate constant k of a reaction varies with temperature according to the equation logk = constant - Ea2.303 \times1T{\text{log}}\,{\text{k}}\,{\text{ = }}\,{\text{constant}}\,{\text{ - }}\,\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303}}}}\,{\text{ \times }}\,\dfrac{{\text{1}}}{{\text{T}}}​ where Ea{E_a} ​ is the energy of activation for the reaction. When a graph is plotted for logk{\text{log}}\,{\text{k}} vs 1T\dfrac{{\text{1}}}{{\text{T}}} a straight line with a slope 6670K - 6670\,K is obtained. The activation energy for this reaction will be: (R=8.314JK1mol1)(R\, = \,8.314\,J{K^{ - 1}}mo{l^{ - 1}})
A.122.65kJmol1A.\,122.65\,kJmo{l^{ - 1}}
B.127.71kJmol1B.\,127.71\,kJmo{l^{ - 1}}
C.142.34kJmol1C.\,142.34\,kJmo{l^{ - 1}}
D.150.00kJmol1D.\,150.00\,kJmo{l^{ - 1}}

Explanation

Solution

The minimum amount of extra energy required by a reacting molecule to get converted into product is called the activation energy. In the presence of a catalyst, the activation energy decreases a bit. The activation energy can also be described as the minimum amount of energy needed to activate molecules or atoms so that they can undergo a chemical reaction. Activation energy is denoted by Ea{E_a} . It is usually measured in joules (J)\left( J \right) and or kJmol1k\,J\,mo{l^{ - 1}} or kcalmol1k\,cal\,mo{l^{ - 1}} .

Complete step-by-step answer: The given Arrhenius equation;
logk = constant - Ea2.303 \times1T{\text{log}}\,{\text{k}}\,{\text{ = }}\,{\text{constant}}\,{\text{ - }}\,\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303}}}}\,{\text{ \times }}\,\dfrac{{\text{1}}}{{\text{T}}}
Let’s write constant as log Alog{\text{ }}A ;
Now, we know that the plotted graph is a straight line; which means,
y=c+mxy\, = \,c\, + \,mx
Therefore, yy will be logk\log \,k, cc will be constant\,{\text{constant}}\,i.e., log Alog{\text{ }}A and the remaining term mxmx will be  - Ea2.303 \times1T{\text{ - }}\,\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303}}}}\,{\text{ \times }}\,\dfrac{{\text{1}}}{{\text{T}}} where xxwill be [1T]\left[ {\dfrac{{\text{1}}}{{\text{T}}}} \right].

In this graph, we are able to observe that [1T]\left[ {\dfrac{{\text{1}}}{{\text{T}}}} \right] which is on xx axis and logk\log \,k values on yy axis.
Where the given intercept value which will be our term cc .
=Ea2.303R= \, - \,\dfrac{{{E_a}}}{{2.303\,R}}
Now, we need to find Ea{E_a}, i.e., activation energy.
So, let’s write the reaction in which the values are given;
slope =6670K = \, - 6670\,K
let’s observe the equation, we need to find activation and we have the value of slope.
So,
Slope of the line =Ea2.303R = \, - \,\dfrac{{{E_a}}}{{2.303\,R}}
Ea=2.303R×slopeoftheline{E_a}\, = \, - 2.303\,R\, \times \,slope\,of\,the\,line
slope=6670Kslope\, = \, - 6670\,K
Ea=2.303R×8.314×(6670){E_a}\, = \, - 2.303\,R\, \times \,8.314\, \times \,( - 6670)
=127711.4= 127711.4
=127.71KJmol1= 127.71\,KJmo{l^{ - 1}}
So, the activation energy is 127.71KJmol1127.71\,KJmo{l^{ - 1}}

Therefore, the correct answer is option B.127.71kJmol1B.\,127.71\,kJmo{l^{ - 1}}

Note: A catalyst is a chemical substance that either increases or decreases the rate of a chemical reaction. In the case of activation energy, a catalyst lowers it. Since, the energies of the reactants will remain the same. A catalyst only changes the activation energy. There can be positive catalysts or else negative catalysts too.