For a reaction,given below is the graph of ln | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.Taking the natural logarithm on both sides,we get $\ln k = -\frac{E_a}{R}(\frac{1}{T}) + \ln

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.Taking the natural logarithm on both sides,we get $\ln k = -\frac{E_a}{R}(\frac{1}{T}) + \ln A$.Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R}$.From the given graph,the slope is calculated as $\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 20}{5 - 0} = -4$.Equating the slopes: $-\frac{E_a}{R} = -4$,which gives $E_a = 4 	imes R$.Given $R = 2 \ cal \ K^{-1} \ mol^{-1}$,we have $E_a = 4 	imes 2 = 8 \ cal \ mol^{-1}$.

For a reaction,given below is the graph of $\ln k$ vs $\frac{1}{T}$. The activation energy for the reaction is equal to $...... \ cal \ mol^{-1}$. (Nearest integer). (Given : $R = 2 \ cal \ K^{-1} \ mol^{-1}$)

Similar Questions

Assertion : If the activation energy of a reaction is zero,temperature will have no effect on the rate constant.
Reason : Lower the activation energy,faster is the reaction.

Which of the following plots gives the value of activation energy?

Slope of the straight line obtained by plotting $\log_{10} k$ against $\frac{1}{T}$ represents which term?

Consider the following graph of the kinetic energy distribution among molecules at temperature $T.$ If the temperature is increased,how would the resulting graph differ from the one above $:-$

Which among the following is correct when energy of activation,$E_a$ of the catalyzed reaction decreases at constant temperature and for same concentration?

Vedclass Test Series

Exam Paper Generator

Online Exam Module