The decomposition of A into product has value of K as 4.5 x 10³ S^–1 at 10°C and energy of activation 60 kJ mol^–1. At what temperature would K be 1.5 x 10⁴ S^–1?

Solution

K₁ = 4.5 x 10³ S^–1 , T₁ = 273 + 10 = 283 K
Ea = 60 kJ mol^–1 T₂ = ?
K₂ = 1.5 x 10⁴S^–1We know that

$\log (\frac{K_{2}}{K_{1}}) = \frac{E_{a} (T_{2} - T_{1})}{2.303 R T_{1} T_{2}} \log (\frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}) = \frac{60 (T_{2} - 283)}{2.303 \times 8.314 \times 283 \times T_{2}} \log (\frac{150}{45}) = \frac{60 (T_{2} \times 283)}{5418.64 T_{2}} \log 150 - \log 45 = \frac{60 T_{2} - 16980}{5418.64 T_{2}} 2.1761 - 1.6532 = \frac{60 T_{2} - 16980}{5418.64 T_{2}} 0.5229 = \frac{60 T_{2} - 16980}{5418.64 T_{2}} 2833.4 T_{2} = 60 T_{2} - 16980 2833.4 T_{2} = 60 T_{2} = - 16980 2773.4 T_{2} = 16980 T_{2} = \frac{16980}{2733.4} = 6.1 K$

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Chemical Kinetics

The decomposition of A into product has value of K as 4.5 x 10³ S^–1 at 10°C and energy of activation 60 kJ mol^–1. At what temperature would K be 1.5 x 10⁴ S^–1?

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The decomposition of A into product has value of K as 4.5 x 10³ S^–1 at 10°C and energy of activation 60 kJ mol^–1. At what temperature would K be 1.5 x 10⁴ S^–1?

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