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GuDViN [60]
3 years ago
13

If a temperature increase from 12.0 ∘c to 20.0 ∘c doubles the rate constant for a reaction, what is the value of the activation

barrier for the reaction?
Chemistry
1 answer:
Blababa [14]3 years ago
7 0
According to this formula:
㏑(K2/K1) = Ea / R / (1/T1 - 1/T2)

Ea = R ㏑(K2/K1) / (1/T1 - 1/T2)
When (K2/K1) the rate constant for the reaction = 2 (because it's doubled) 
and we have R constant = 8.314 and T1= 12 + 273 = 285 kelvin T2 = 20 + 273=293 Kelvin
So by substitution:
∴ Ea (the activation barrier for the reaction) =( 8.314* ㏑(2) ) / (1/285 - 1/293)
              = 60155 J Mol^-1 = 60 KJmol^-1
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7 0
3 years ago
Complete and balance the following redox reaction in basic solution
inn [45]

Answer:

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Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Answer

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Explanation:

Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)

add H^1+ (acid) to capture the O and make 7 water molecules

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3 Hg on the right and left

14 H on the right and left

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