The rate constant for the first-order reaction is given by log10 k = 14.34 – 1.25 × 104 T. Calculate activation energy of the reaction.

The rate constant for the first-order reaction is given by log₁₀ k = 14.34 – 1.25 × 10⁴ T. Calculate activation energy of the reaction.

Sum

The given rate constant equation is

log₁₀k = 14.34 – 1.25 × 10⁴ T ...(1)

Arrhenius equation is

k = `"Ae"^((-"E"_"a")/("RT"))`

∴ In k = InA - `"E"_"a"/("RT")`

∴ log₁₀k = log₁₀A - `"E"_"a"/(2.303 "RT")` ...(2)

Comparing (1) and (2),

`"E"_"a"/(2.303 "RT") = 1.25 xx 10^4 "T"`

∴ `"E"_"a"/(2.303 "R") = 1.25 xx 10^4`

∴ `"E"_"a"/(2.303 xx 8.314) = 1.25 xx 10^4`

∴ E_a = 1.25 × 10⁴ × 2.303 × R

∴ E_a = 1.25 × 10⁴ × 2.303 × 8.314

= 23.93 × 10⁴J

= 239.3 kJ mol^-1

The energy of activation of the reaction is 239.3 kJ mol^-1

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Chapter 6: Chemical Kinetics - Exercises [Page 137]

Chapter 6 Chemical Kinetics
Exercises | Q 4. viii. | Page 137

Answer the following in one or two sentences.

Write the Arrhenius equation and explain the terms involved in it.

Explain with the help of the Arrhenius equation, how do the rate of reaction changes with temperature.

Answer the following in brief.

How will you determine activation energy graphically using the Arrhenius equation?

Answer the following in brief.

Explain graphically the effect of temperature on the rate of reaction.

The rate constant of a reaction at 500°C is 1.6 × 10³ M⁻¹ s⁻¹. What is the frequency factor of the reaction if its activation energy is 56 kJ/mol?

Solve

What fraction of molecules in a gas at 300 K collide with an energy equal to the activation energy of 50 kJ/mol?

Which among the following equation represents Arrhenius equation?

Slope of the straight line obtained by plotting log₁₀k against represents what term?