The rate of the chemical reaction doubles for an increase of 10 k in absolute temperature from 298 K. Calculate E_a.

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Solution

Given that initial temperature, T₁ = 298 K
Final temperature, T₂ = 298 + 10 = 308 K
Rate of chemical reaction will increase with raise of rate constant so that take
Initial value of rate constant k₁ = k
find rate constant, k₂ = 2k
Also, R = 8.314 J K⁻¹ mol⁻¹
Use Arrhenius equation

According to Arrchenium equation,

\log \frac{k_{2}}{k_{1}} = \frac{Ea}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]

\frac{k_{2}}{k_{1}} = 2, T_{1} = 298 K, T_{2} = 298 + 10 = 308 k R = 8.314 J k^{- 1} {mol}^{- 1}

\log 2 = \frac{Ea}{2.303 \times (8.314 J K^{- 1} {mol}^{- 1})} [\frac{1}{298 k} - \frac{1}{308 k}]

0.3010 = \frac{Ea}{2.303 \times 8.314} \times \frac{10}{298 \times 308}

Ea = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10} J {mol}^{- 1}

Ea = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10 \times 100} kJ {mol}^{- 1} = 52.89 kJ {mol}^{- 1}

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The rate of the chemical reaction doubles for an increase of 10 k in absolute temperature from 298 K. Calculate E_a.

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For Daily Free Study Material Join wiredfaculty

Chemical Kinetics

The rate of the chemical reaction doubles for an increase of 10 k in absolute temperature from 298 K. Calculate Ea.

Some More Questions From Chemical Kinetics Chapter

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Mock Test Series

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The rate of the chemical reaction doubles for an increase of 10 k in absolute temperature from 298 K. Calculate E_a.

The rate law for the decomposition of N₂O₅ is rate = k[N₂O₅]. What is the significance of k in this equation?