Q

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

4.28  The decomposition of A into product has value of k as $4.5 \times 10 ^3 s ^{-1}$ at 10°Cand energy of activation 60 kJ mol–1. At what temperature would k be $1.5 \times 10 ^4 s ^{-1}$ ?

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The decomposition of A into a product has a value of k as $4.5 \times 10 ^3 s ^{-1}$ at 10°C and energy of activation 60 kJ mol–1.

K1 = $4.5 \times 10 ^3 s ^{-1}$
K2 = $1.5 \times 10 ^4 s ^{-1}$

$E_a$ =  60 kJ mol–1

K2 = $1.5 \times 10 ^4 s ^{-1}$

$log\frac{K_2}{K_1}=\frac{E_a(T_2-T_1)}{2.303RT_1T_2}$

$log(\frac{1.5\times 10^{4}}{4.5\times 10^{3}})=\frac{60(T_2-283)}{2.303\times R\times 283\times T_2}$

$log(\frac{150}{45})=\frac{60(T_2-283)}{5418.61\times T_2}$

$log150-log45=\frac{60T_2-16980}{5418.61\times T_2}$

$2.176-1.653=\frac{60T_2-16980}{5418.61\times T_2}$

$0.5229=\frac{60T_2-16980}{5418.61\times T_2}$

$T_2=\frac{16980}{2733.4}$

$T_2=6.1K$

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