Question

At 380°C, the half-life period for the first order decomposition of H₂O₂ is 360 minutes. The energy of activation of the reaction is 200 kJ mol⁻¹. Calculate the time required for 75% decomposition at 450°C.

Answer 1

Given, `t_(1//2)^(380^circ C)` = 360 minutes, E_a = 200 kJ mol⁻¹

∴ k^380°C = `0.693/(t_(1//2)^(380^circ C))`

= `0.693/360`

= 1.925 × 10⁻³ min⁻¹

According to the Arrhenius equation,

`log_10  k_2/k_1 = E_a/(2.303 R) [1/T_1 - 1/T_2]`

Putting T₁ = 380 + 273 = 653 K and 

T₂ = 450 + 273 = 723 K, we have

`log_10  (k^(450^circ C))/(k^(380^circ C)) = E_a/(2.303 R) [1/T_1 - 1/T_2]`

or, `log_10  (k^(450^circ C))/(1.925 xx 10^-3) = 200/(2.303 xx 8.314 xx 10^-3) [(723 - 653)/(653 xx 723)]`    ...(∵ R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹)

or, `log_10  (k^(450^circ C))/(1.925 xx 10^-3) = 200/0.01914 [70/472119]`

`log_10  (k^(450^circ C))/(1.925 xx 10^-3)` = 10449.320 × 1.4826 × 10⁻⁴

`log_10  (k^(450^circ C))/(1.925 xx 10^-3)` = 1.549

or, k^450°C = 1.925 × 10⁻³ × antilog₁₀ 1.549

= 6.814 × 10⁻² min⁻¹

For a first order reaction, t = `2.303/k log_10  ([A]_0)/([A])`

For 75% decomposition,

[A] = `[A]_0 - [A]_0 xx 75/100`

= [A]₀ × 0.25

∴ t = `2.303/(6.814 xx 10^-2) log_10  ([A]_0)/([A]_0 xx 0.25)`

= 20.35 minutes.

Hence, the time required for 75% decomposition at 450°C is 20.35 minutes.