Question

Temperature coefficient, $\mu \quad =\quad \cfrac { { k }_{ { 35 }^{ 0 }C } }{ { k }_{ { 25 }^{ 0 }C } }$ of a reaction is 1.82. Calculate the energy of activation in calories. $(R = 1.987 cal {degree}^{-1}{mol}^{-1})$.

Answer 1

Hint: Activation energy can be defined as the minimum amount of extra energy that is required by a reacting molecule to get converted into a product. It can also be defined as the minimum amount of energy that is needed to activate or energize the molecules or atoms so that they can undergo a chemical reaction or transformation.

Complete step by step answer: It is given in the question that the value of the temperature coefficient which is the ratio of the equilibrium constant at ${25}^{0}C$ and equilibrium constant at ${35}^{0}C$ is 1.82 and we need to find out the energy of activation, ${E}_{a}$ in calories.

It is given that $\mu \quad =\quad \cfrac { { k }_{ { 35 }^{ 0 }C } }{ { k }_{ { 25 }^{ 0 }C } } \quad = \quad 1.82$.

And from Arrhenius equation, we have,
$\cfrac { { k }_{ { 35 }^{ 0 }C } }{ { k }_{ { 25 }^{ 0 }C } } \quad =\quad \cfrac { { E }_{ a } }{ 2.303R } \left[ \cfrac { 1 }{ { T }_{ 1 } } \quad -\quad \cfrac { 1 }{ { T }_{ 2 } } \right]$

Replacing $\cfrac { { k }_{ { 35 }^{ 0 }C } }{ { k }_{ { 25 }^{ 0 }C } }$ with $\mu$, we get
$\implies \mu \quad =\quad \cfrac { { E }_{ a } }{ 2.303R } \left[ \cfrac { 1 }{ { T }_{ 1 } } \quad -\quad \cfrac { 1 }{ { T }_{ 2 } } \right]$
Where, ${E}_{a} = Activation \quad energy$ ${T}_{1} = {25}^{0}C$, ${T}_{2} = {35}^{0}C$ and $R = 1.987 \quad cal \quad {degree}^{-1} \quad {mol}^{-1}$.

Substituting these values in equation (1), we get,
$1.82\quad =\quad \cfrac { { E }_{ a } }{ 2.303\quad \times \quad 1.987 } \left[ \cfrac { 1 }{ 25 } \quad -\quad \cfrac { 1 }{ 35 } \right]$

Taking $2.303 \quad \times \quad 1.987$ to the left hand side of the equation, we get
$\implies1.82\quad \times \quad 2.303\quad \times \quad 1.987\quad =\quad { E }_{ a }\left[ \cfrac { 10 }{ 875 } \right]$

Now, solving for ${E}_{a}$, we get
$\implies{ E }_{ a }\quad =\quad 1.82\quad \times \quad 2.303\quad \times \quad 1.987\quad \times \quad 875\quad \times \quad 0.1$
$\implies{ E }_{ a }\quad =\quad 728.7\quad cal/mol$

Therefore, the activation energy is 728.7 cal/mol.

Note: Energy is defined as the capacity for some work whereas activation energy is the energy needed to form an activated complex during the chemical reaction. Activation energy is a type of energy needed to initiate or activate a reaction.