Skip to content
Philoid
Browse Saved
Back to chapter
Chemistry
4. Chemical Kinetics
Home · Class 12 · Chemistry · Chemistry Part-I · 4. Chemical Kinetics
Prev
Next
Q26 of 39 Page 117

The decomposition of hydrocarbon follows the equation

k = (4.5 × 1011s–1) e-28000K/T. Calculate Ea.

The given equation is


k = (4.5 x 1011 s-1) e-28000 K/T (i)


Comparing, Arrhenius equation


k = Ae -Ea/RT (ii)


We get, Ea / RT = 28000K / T


⇒ Ea = R x 28000K


= 8.314 J K-1mol-1 × 28000 K


= 232792 J mol–1 or 232.792 kJ mol–1


More from this chapter

All 39 →
24

Consider a certain reaction A → Products with k = 2.0 × 10–2s-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L–1.

25

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

27

The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T

Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

28

The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104s–1?

Questions · 39
4. Chemical Kinetics
1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Back to chapter
ADVERTISEMENT
About Contact Privacy Terms
Philoid · 2026
  • Home
  • Search
  • Browse
  • Quiz
  • Saved