Question

The rate constant for the first order decomposition of a certain reaction is given by the equation

ln k (s⁻¹) = `14.34 - (1.25 xx 10^4)/T`

Calculate

1. the energy of activation,
2. the rate constant at 500 K.
3. At what temperature will its half-life period be 256 minutes?

Answer 1

The given equation is

ln k (s⁻¹) = `14.34 - (1.25 xx 10^4)/T`    ...(i)

According to the Arrhenius equation,

k = `A e^(-E_a/(RT)`

Taking logarithm, we get

ln k = `ln A - E_a/(RT)`    ...(ii)

a. Comparing eqs. (i) and (ii), we get

`E_a/R` = 1.25 × 10⁴

or, E_a = 1.25 × 10⁴ × R

= 1.25 × 10⁴ × 8.314

= 103925 J mol⁻¹

= 103.925 kJ mol⁻¹

b. Putting.T = 500 K in eq. (i), we get

ln k = `14.34 - (1.25 xx 10^4)/500`

or, 2.303 log₁₀ = `14.34 - (1.25 xx 10^4)/500`

= −10.66

or, k = `"antilog"_10 (-10.66)/2.303`

= 2.35 × 10⁻⁵ s⁻¹

c. The value of k corresponding to t_1/2 = 256 min = 256 × 60 s is given by

k = `0.693/t_(1//2)`

= `0.693/(256 xx 60)`

= 4.51 × 10⁻⁵ s⁻¹

Putting this value of k in eq. (i), we get

ln 4.51 × 10⁻⁵ = `14.34 - (1.25 xx 10^4)/T`

or, 2.303 × log₁₀ 4.51 × 10⁻⁵ = `14.34 - (1.25 xx 10^4)/T`

or, `(1.25 xx 10^4)/T` = 14.34 −  2.303 × log₁₀ 4.51 × 10⁻⁵

⇒ `(1.25 xx 10^4)/T` = 24.348

or, `T = (1.25 xx 10^4)/24.348`

T = 513.4

Hence, at 513.4 K, the reaction will have a half-life of 256 minutes.