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Crazy boy [7]
2 years ago
10

A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =

85.6 kj/mol?
Chemistry
1 answer:
IgorC [24]2 years ago
8 0
According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
  = 2.16 x10^2
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3 years ago
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Andrej [43]

Given parameters:

Mass of KClO₃  = 125g

Unknown:

Total mass of the products = ?

When  KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

               2KClO₃  →  2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

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  molar mass of  KClO₃  = 39 + 35.5 + 3(16)  = 122.5g/mol

    So number of moles of KClO₃ = \frac{125}{122.5}  = 1.02moles

    2. Now, using this number of moles, find the number of moles of the products using this value;

   2 moles of KClO₃ produced 2 moles of KCl

  1.02 moles of KClO₃ will also produce 1.02moles of KCl

   2 moles of KClO₃ produced 3 moles of O₂

   1.02 moles of KClO₃ will produce   \frac{1.02 x 3} {2} mole = 1.53 moles of O₂

   3. Now find the masses of each product;

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Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

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