Answer
Verified
387.6k+ views
Hint: First take the Arrhenius equation and do logs on both sides. Then to the obtained equation put limits of T tends to ∞. This will lead you to your answer.
Complete step by step answer:
- First we need to see what the Arrhenius equation is.
It was experimentally proved that with rise in temperature by ${10^ \circ }$, the rate constantly gets doubled of its initial value. So, the Arrhenius equation was formulated to accurately explain the temperature dependence of the rate of a chemical reaction.
$k = A{e^{\frac{{ - {E_a}}}{{RT}}}}$
Or by taking natural logarithms on both sides: $\ln k = \ln A - \frac{{{E_a}}}{{RT}}$
Where A = Arrhenius factor or frequency factor or pre exponential factor;
R = gas constant;
${E_a}$ = activation energy
- Above equation can also be written as:
$\log k = \log A - \frac{{{E_a}}}{{2.303RT}}$
Now we will be taking limits on both sides:
$\mathop {\lim }\limits_{T \to \infty } (\log k) = \mathop {\lim }\limits_{T \to \infty } \left[ {\log A - \frac{{{E_a}}}{{RT}}} \right]$
Since there is no T variable in term logA so the limit will not apply to that. The equation now becomes:
$\mathop {\lim }\limits_{T \to \infty } (\log k) = \log A - \mathop {\lim }\limits_{T \to \infty } \left[ {\frac{{{E_a}}}{{RT}}} \right]$
Now putting the values: we know that if T = ∞, $\frac{1}{T}$ will be = 0.
And so: $\frac{{{E_a}}}{{RT}}$ term will be = 0.
So, the equation becomes: $\mathop {\lim }\limits_{T \to \infty } (\log k) = \log A$
So, the correct option will be: (D) logA.
Note:According to the Arrhenius equation, the lower the activation energy faster will be the rate of the reaction. Even a small amount of catalyst can catalyse a large number of reactants and so to increase the rate of a reaction a catalyst can be added which provides an alternate pathway for the reaction by reducing the activation energy and thus reducing the potential energy barrier.
Complete step by step answer:
- First we need to see what the Arrhenius equation is.
It was experimentally proved that with rise in temperature by ${10^ \circ }$, the rate constantly gets doubled of its initial value. So, the Arrhenius equation was formulated to accurately explain the temperature dependence of the rate of a chemical reaction.
$k = A{e^{\frac{{ - {E_a}}}{{RT}}}}$
Or by taking natural logarithms on both sides: $\ln k = \ln A - \frac{{{E_a}}}{{RT}}$
Where A = Arrhenius factor or frequency factor or pre exponential factor;
R = gas constant;
${E_a}$ = activation energy
- Above equation can also be written as:
$\log k = \log A - \frac{{{E_a}}}{{2.303RT}}$
Now we will be taking limits on both sides:
$\mathop {\lim }\limits_{T \to \infty } (\log k) = \mathop {\lim }\limits_{T \to \infty } \left[ {\log A - \frac{{{E_a}}}{{RT}}} \right]$
Since there is no T variable in term logA so the limit will not apply to that. The equation now becomes:
$\mathop {\lim }\limits_{T \to \infty } (\log k) = \log A - \mathop {\lim }\limits_{T \to \infty } \left[ {\frac{{{E_a}}}{{RT}}} \right]$
Now putting the values: we know that if T = ∞, $\frac{1}{T}$ will be = 0.
And so: $\frac{{{E_a}}}{{RT}}$ term will be = 0.
So, the equation becomes: $\mathop {\lim }\limits_{T \to \infty } (\log k) = \log A$
So, the correct option will be: (D) logA.
Note:According to the Arrhenius equation, the lower the activation energy faster will be the rate of the reaction. Even a small amount of catalyst can catalyse a large number of reactants and so to increase the rate of a reaction a catalyst can be added which provides an alternate pathway for the reaction by reducing the activation energy and thus reducing the potential energy barrier.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE