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kramer
3 years ago
5

A certain 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 h

igher temperature to f at the lower temperature?
Chemistry
1 answer:
lapo4ka [179]3 years ago
8 0
Arrhenius' Law relates activation energy, Ea, rate constant, K, and temperature, T as per this equation:

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

Answer: 7.5




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So you know you have 5.4 g of NaCl and you need to know how many moles there are in this amount of NaCl

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<u>Written-out method:</u>

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<u>Plug into calculator method:</u>

(5.4 g of NaCl/ 58.44g NaCl= .0925 moles)

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