Question

Using Microsoft excel or any other plotting softwares, calculate the activation energy from the following table. (a) Use all data. (b) Use only 25°C and 30°C. (c) Compare the two values. Are they different? Why? 4. Consider the following hypothetical mechanism: Step 1 N2O (g) -->N2 (g) + O (g) slow Step 2 N2O (g) + O (g) --> N2 (g) + O2 (g) fast What is the overall reaction? What is the rate law of the reaction if the first step is the slow step (rate determining step or RDS)? T emperature(°C) 25 30 35 40 45 50 Rate constant (M/s) 8.8x10-5 1.6x10-4 2.8x10-4 5.0x10-4 8.5x10-4 1.4x10-3 k1 k2 What is the molecularity of the reaction if the first step is the slow step? What are the intermediate species?

Answer 1

Ans 1

Part a

From Arrhenius equation

ln (k2/k1) = (Ea/R) (1/T1 - 1/T2)

Y = mX

Y = ln(k2/k1)

Slope m = (Ea/R)

X = (1/T1 - 1/T2)

Make Arrhenius plot

Slope = - Ea/R = - 10658

Ea = 10658 x 8.314 = 88610.612 J/mol

Part b

T1 = 25 + 273 = 298 K

T2 = 30 + 273 = 303 K

Rate constant k1 = 8.8*10^-5 M/s

Rate constant k2 = 1.6*10^-4 M/s

Gas constant R = 8.314 J/mol·K

From Arrhenius equation

ln (k2/k1) = (Ea/R) (1/T1 - 1/T2)

ln (1.6*10^-4/8.8*10^-5) = (Ea/8.314) (1/298 - 1/303)

0.59779 = 6.66 x 10^-6 x Ea

Ea = 89752.71 J/mol

Part c

% error = (89752.71 - 88610.612) *100 / (88610.612)

= 1.29%

If we use two data points instead of all data points simultaneously then we get 1.29% error in the result.