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# The decomposition of A into product has value of k as 4.5 x 10^3 s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 x 10^4 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?

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Arrhenius equation, we know that,

log $$\frac{k_2}{k_1}$$ = $$\frac{E_a}{2.303R}$$ $$\Big(\frac{T_2 - T_1}{T_1T_2}\Big)$$

Also, k1 = 4.5 × 103 s-1

T1 = 273 + 10 = 283 K

k2 = 1.5 × 104 s-1

Ea = 60 kJ mol-1 = 6.0 × 104 J mol-1

Then,

log $$\frac{1.5 \times 10^4}{4.5 \times 10^3}$$

= $$\frac{6.0 \times 10^4 \ J mol^{-1}}{2.303 \times 8.314 \ J \ K^{-1} mol^{-1}}$$$$\Big(\frac{T_2 - 283}{283T_2}\Big)$$

⇒ 0.5229 = 3133.627$$\Big(\frac{T_2 - 283}{283T_2}\Big)$$

⇒ $$\frac{0.5229 \times 283 T_2}{3133.627}$$ = T2 - 283

⇒ 0.0472 T2 = T2 - 283

⇒ 0.0472 T2 = 283

⇒ T2 = 297.019 K(approximately)

= 297 K

= 24°C

Hence, k would be 1.5 × 104 s-1 at 24°C.

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