The activation energy of a reaction can be determined from the slope of which of the following graphs?

Question

The activation energy of a reaction can be determined from the slope of which of the following graphs? (A) In K vs. T (B) In K vs. (1/T) (C) < (I n K/T) vs. (1/T) (D) (I n K/T) vs. T

Tardigrade Admin · Accepted Answer

Arrhenius equation K=A.e- E a / R T lnK=lnA-(E a/R T) So, activation energy of reaction can be determined from the slope of lnK vs (1/T) . <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-hiz47idm8q2d.png" /> Using y=mx+c we can find the value of Ea from slope value.

Q. The activation energy of a reaction can be determined from the slope of which of the following graphs?

In K vs. T

In K vs. $\frac{1}{T}$

< $\frac{I n K}{T}$ vs. $\frac{1}{T}$

$\frac{I n K}{T}$ vs. T

Arrhenius equation $K = A . e^{- E a / RT}$

$l n K = l n A - \frac{E a}{RT}$

So, activation energy of reaction can be determined from the slope of lnK vs $\frac{1}{T}$ .

Using $y = m x + c$ we can find the value of $E_{a}$ from slope value.

Q. The activation energy of a reaction can be determined from the slope of which of the following graphs?

In K vs. T

In K vs. T1​

< TInK​ vs. T1​

TInK​ vs. T

Arrhenius equation K=A.e−Ea/RT lnK=lnA−RTEa​ So, activation energy of reaction can be determined from the slope of lnK vs T1​ . Using y=mx+c we can find the value of Ea​ from slope value.

In K vs. $\frac{1}{T}$

< $\frac{I n K}{T}$ vs. $\frac{1}{T}$

$\frac{I n K}{T}$ vs. T

Arrhenius equation $K = A . e^{- E a / RT}$

$l n K = l n A - \frac{E a}{RT}$

So, activation energy of reaction can be determined from the slope of lnK vs $\frac{1}{T}$ .

Using $y = m x + c$ we can find the value of $E_{a}$ from slope value.