The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 - 1.25 x 104 K/T.
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Given, Arrhenius equation is,
k = Ae-Ea/RT
⇒ ln k = ln A -\(\frac{E_a}{RT}\)
⇒ ln k = log A -\(\frac{E_a}{RT}\)
⇒ log k = log A -\(\frac{E_a}{2.303 \ RT}\) ....(i)
The given equation is
log k = 14.34 - 1.25 x 104 K/T ....(ii)
From equation (i) and (ii), we obtain
\(\frac{E_a}{2.303 \ RT}\) = \(\frac{1.25 \times 10^4 K}{T}\)
⇒ Ea = 1.25 x 104 K x 2.303 x R
= 1.25 × 104 K × 2.303 × 8.314 J K-1 mol-1
= 239339.3 J mol-1 (approximately)
= 239.34 kJ mol-1
Also, when t1/2 = 256 minutes,
k = \(\frac{0.693}{t_{1/2}}\)
= \(\frac{0.693}{256}\)
= 2.707 × 10-3 min-1
= 4.51 × 10-5 s-1
It is also given that,
log k = 14.34 - 1.25 × 104 K/T
⇒ log(4.51 x 10-5) = 14.34 - \(\frac{1.25 \times 10^4 \ K}{T}\)
⇒ log(0.654 - 05) = 14.34 - \(\frac{1.25 \times 10^4 \ K}{T}\)
⇒ \(\frac{1.25 \times 10^4 \ K}{T}\) = 18.686
⇒ T = \(\frac{1.25 \times 10^4 \ K}{18.686}\)
= 668.95 K
= 669 K (approximately)