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Veronika [31]
1 year ago
11

A first order reaction has a rate constant of 0. 543 at 25°C. Given that the activation energy is 75. 9 kj/mol. Calculate the ra

te constant at 32. 3 °C.
Chemistry
1 answer:
Rasek [7]1 year ago
4 0

The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.

First-order reactions are very commonplace. we have already encountered  examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to present t-butanol. every other reaction that famous obvious first-order kinetics is the hydrolysis of the anticancer drug cisplatin.

The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.

Rate constant K₁  = 0. 543 /s

T₁  = 25°C

Activation energy Eₐ =  75. 9 k j/mol.

T₂ = 32. 3 °C.

K₂ =?

formula;

log K₂/K₁=  Eₐ /2.303 R [1/T₁ - 1/T₂]

putting the value in the equation  

K₂ = 0.343 /s

Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s

The specific rate steady is the proportionality consistent touching on the fee of the reaction to the concentrations of reactants. The fee law and the specific charge consistent for any chemical reaction should be determined experimentally. The cost of the charge steady is temperature established.

Learn more about activation energy here:- brainly.com/question/26724488

#SPJ4

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How many moles of h2 can be formed if a 3.25g sample of Mg reacts with excess HCl
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0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

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The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:

  • Mg: 1 mole
  • HCl: 2 moles
  • MgCl₂: 1 mole
  • H₂: 1 mole

Being:

  • Mg: 24. 31 g/mole
  • H: 1 g/mole
  • Cl: 35.45 g/mole

the molar mass of the compounds participating in the reaction is:

  • Mg: 24.31 g/mole
  • HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
  • MgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/mole
  • H₂: 2*1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:

  • Mg: 1 mole* 24.31 g/mole= 24.31 g
  • HCl: 2 moles* 36.45 g/mole= 72.9 g
  • MgCl₂: 1 mole* 95.21 g/mole= 95.21 g
  • H₂: 1 mole* 2 g/mole= 2 g

Then you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?

moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}

moles of H₂= 0.134

<u><em>0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl</em></u>

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3 years ago
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