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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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A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr
sertanlavr [38]

Answer: The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.947 atm

P_2 = final pressure of gas = 0.987 atm

V_1 = initial volume of gas = 150 ml

V_2 = final volume of gas = 144 ml    

T_1 = initial temperature of gas = 25^0C=(25+273.15)K=298.15K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}

T_2=298.31K=(298.31-273.15)^0C=25.16^0C

The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

4 0
2 years ago
Classify the following substituents according to whether they are electron donors or electron acceptors relative to hydrogen by
igor_vitrenko [27]

Species that have a lone pair of electrons often donate electrons by resonance while substituents that are electron deficient take away electrons by resonance.

<h3>What is resonance?</h3>

The term resonace has to do with the movement of electron pairs in a molecule. Inductive effects has to do with the drawing of electron density towards an atom or bond.

The two effects depends on the nature of a substituent. For instance, species that have a lone pair of electrons often donate electrons by resonance while substituents that are electron deficient take away electrons by resonance.

The question is incomplete hence the exact nature of the substituents can not be determined.

Learn more about resonance: brainly.com/question/23287285?

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(WILL GIVE BRAINLIEST) A student uses res litmus paper to test the pH value of different solutions. Which would result in a colo
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Answer:

Soap & Drain Cleaner

Explanation:

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The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p
earnstyle [38]

Answer:

The frequency is  f =  2,01 * 10^{15} \  Hz

Explanation:

From the question we are told that

   The energy required to ionize boron is E_b  =  801 KJ/mol

Generally the ionization energy of boron pre atom is mathematically represented as

     E_a  =  \frac{E_b}{N_A}

Here  N_A is the Avogadro's constant with value N_A  =  6.022*10^{23}

So

      E_a  =  \frac{801}{6.022*10^{23}}

=>     E_a  =  1.330 *10^{-18} \  J/atom

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

       E =  hf  =  E_a

=>     hf  =  E_a

Here h is the Planks constant with value h = 6.626 *10^{-34}

So

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Using the concepts of ellipse and the universal law of gravitation it can be found that the correct answer is:

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In the case of the sun and the planet, the sun is many orders of magnitude more massive than the planets, therefore, it is almost fixed in the center and the lighter planets rotate around it, where the sun is in the focus of the ellipse orbit.

In the special case that the two foci coincide, there is a circular movement, but the most general is that there is some separation between the foci, so the movement is elliptical with the sun located in one of the foci.

Let's analyze the claims of this exercise

a) True. When the orbit is not a perfect circle, you have an elliptical orbit, this is the most common case.

b) False. The axis of the planet has no effect on the orbit, it affects the intensity of the radiation that reaches the planets

c) False. Circular motion is a special case of elliptical motion when the two foci coincide

d) False. The axis of the planet influences the intensity of radiation reaching the surface, but not the shape of the orbit of the plant.

Using the concepts of ellipse it can be found that the correct answer is:

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