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The activation energy of the reaction: A...

The activation energy of the reaction: `A + B rarr` Products is `105.73 kJ mol^(-1)`. At `40^(@)C`, the Products are found at the rate of `0.133 mol L^(-1) min^(-1)`. What will be the rate of formation of Products at `80^(@)C`?

Text Solution

Verified by Experts

Let the rate law be defined as
At `T_1 : r_1 = k_1 [A]^x [B]^y`
At ` T_2: r_2 = k_2 [A]^x [B]^y rArr r_2= r_1(k_2/k_1)`
Using Arrhenius equation, find k at `40^@C`.
`log_10 k_2/k_1 = E_a/(2.303 R) ((T_2-T_1)/(T_1T_2)) rArr log_10 k_2/k_1= (105.73 xx 10^(3))/(2.303 xx 8.31)(40/(313 xx 353))`
`rArr log_10 k_2/ k_1 = 2.0 rArr k_2/k_1 = 100 rArr r_2 = 0.133 xx 100 = 13.3 mol/ L/min`
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