The rate constant for the decomposition of N2O5 at various temperatures is given below:


T/°C



0



20



40



60



80



105 × k/s-1



0.0787



1.70



25.7



178



2140



Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.


To convert the temperature in °C to °K we add 273 K.


 



 



The graph is given as:


The Arrhenius equation is given by k = Ae-Ea/RT


Where, k- Rate constant


A- Constant


Ea-Activation Energy


 


R- Gas constant


 


 


T-Temperature


 


Taking natural log on both sides,


ln k = ln A-(Ea/RT)


 


→ equation 1


 


By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –Ea/R.


Slope = (y2-y1)/(x2-x1)


By substituting the values, slope = -12.301


⇒ –Ea/R = -12.301


But, R = 8.314 JK-1mol-1


⇒ Ea = 8.314 JK-1mol-1 × 12.301 K


⇒ Ea = 102.27 kJ mol-1


Substituting the values in equation 1 for data at T = 273K


 



 



 


(∵ At T = 273K, ln k = -7.147)


 


On solving, we get ln A = 37.911


∴ A = 2.91×106


 


When T = 300C, hence T = 30 + 273 = 303K


 




 


⇒ ln k = -2.8


 


⇒ k = 6.08x10-2s-1.


 


When T = 500C, hence T = 50 + 273 = 323K


 


Substituting in equation 1,


 A = 2.91 × 106, Ea = 102.27 kJ mol-1


 


Ln K = ln(2.91 × 106)- 102.27/(8.314 × 323)


 


Thus, ln k = -0.5 and k = 0.607s-1.


 


 


 


 


 


 


 


 

22
1