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twrxght6643
01.09.2019 •
Chemistry
Acertain reaction with an activation energy of 205 kj/mol was run at 485 k and again at 505 k . what is the ratio of f at the higher temperature to f at the lower temperature?
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Ответ:
K (T) = A * e ^ (-Ea / RT), where R is the universal constant of gases and A is a constant which accounts for collision frequency..
Then you can find the ration between K's at two different temperatures as:
K1 = A * e ^ (-Ea / RT1)
K2 = A* e ^(-Ea / RT2)
=> K1 / K2 = e ^ { (-Ea / RT1) - Ea / RT2) }
=> K1 / K2 = e ^ {(-Ea/ R ) *( 1 / T1 - 1 T2) }
=> K1 / K2 = e^ { (-205,000 j/mol / 8.314 j/mol*k )* ( 1 / 505K - 1/ 485K) }
=> K1 / K2 = e ^ (2.0134494) ≈ 7.5
7.5
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