Therate of reaction depends on the number of collisions between the reactant atoms and molecules, on the fraction of the collisions that occur with proper orientation, and on the fraction that overcomes Ea (activation energy). These are summarized by the Arrhenius equation.

k = A e ^{–Ea/RT}

where:

A is the frequency of collisions with correct geometry;

e ^{–Ea/RT} is the fraction of molecules with minimum energy for reaction at temperature T.

As the temperature increases, the value of A increases due to an increase in the number of collisions. In addition, e ^{–Ea/RT} increases, a reflection of the increase in the fraction of colliding molecules that overcome the activation barrier.

I like to share this Ph Equation with you all through my article.

TheArrhenius equation is used to calculate the value of ‘k’ at various temperatures and also the activation energy of a reaction.

arrhenius equation : Linear relationship:

The Arrhenius equation can be rewritten in the form of a linear relationship.

ln k = ln A – E_{a}/R [1/T]

In k = y

In A = a

-E_{a}/R = b

1/T = x

A plot versus of ln k versus 1/T yields a straight line with a slope of –E_{a}/R.

The slope is Δ ln (k)/ Δ(1/T)

The value of E_{a} (activation energy) can be determined by solving the following equation:

Slope = -E_{a}/R

E_{a} = -R × (slope)

OR

Ea = - R [Δ ln (k)/ Δ (1/T)]

arrhenius equation : Determining Activation Energy:

The reaction

2N_{2}O_{5 (g)} → 4NO_{2 (g)} + O_{2 (g)}

was studied at several temperatures, and the following values of k were obtained:

Calculate the value of E_{a} for this reaction.

*Solution:*

Inorder to solve this problem, we need to plot a graph (In k versus 1/T).Before that we need to find out the values of ln(k), T in Kelvin scale and then 1/T.

Let us now plot the graph:

From the graph Δ(1/T)= (3.365-3.05) × 10 ^{-3}

= 0.315 × 10 ^{-3}

And

Δ(ln k) = -10-(-6.5)

= -10+ 6.5

= -3.5

Chemistry is widely used in day to day activities watch out for my forthcoming posts on Nonmetals on Periodic Table and Periodic Table Au. I am sure they will be helpful.

As we know the slope gives the value for -E_{a}/R

So, E_{a} = -R × (slope)

E_{a}= - (8.314 JK-1mol-1) (Δ(ln k)/ Δ(1/T)

= -(8.314)(-3.5/(0.315 × 10 ^{-3}))

= (-8.314) (-11111.11)

= 92377.777

= 0.92377 x 10 ^{5}

= approximately 1.0 x 10 ^{5} J mol^{-1}