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The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 - 1.25 x 10^4 K/T.

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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?

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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)

This post was modified 8 months ago by admin
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