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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 × 10 ^ 4K/T

4.27   The rate constant for the first order decomposition of H_2O_2 is given by the following equation:
           \log k = 14.34 - 1.25 \times 10 ^ 4K/T .

Calculate E_a for this reaction and at what temperature will its half-period be 256 minutes?

Answers (1)
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M manish

The Arrhenius equation is given by 
k=Ae^{-E_{a}/RT}

taking log on both sides, 

\log k = \log A -\frac{E_{a}}{2.303RT}....................(i)
given equation,

\log k = 14.34 - 1.25 \times 10 ^ 4K/T.....................(ii)

On comparing both equation we get,

E_{a}/2.303R=1.25 \times 10^{4}

activation energy 
\\=1.25 \times 10^{4} \times 2.303 \times 8.314\\ =239.34\ KJ/mol

half life (t_{1/2})  = 256 min

k = 0.693/256 
 k = 4.51\times 10^{-5} s^{-1}

With the help of equation (ii),

\log4.51\times 10^{-5} s^{-1} = 14.34-\frac{1.25\times 10^{4}}{T}

\frac{1.25\times 10^{4}}{T} = 18.686
           T = \frac{1.25\times 10^{4}}{18.686}

               = 669 (approx)

 

 

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