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poizon [28]
9 months ago
12

According to le châtelier’s principle, the amount of solid reactant or product present does not have an impact on the equilibriu

m. Why?.
Chemistry
1 answer:
Inessa05 [86]9 months ago
5 0

The concentration of solids is constant and usually taken equal to unity ,therefore it does not appear in the equilibrium constant ,so adding or removing solid has no effect. So According to Le Chatelet's Principle the amount of solid reactant or product present does not have an impact on the equilibrium

What is  Le Chatelet's Principle ?

The position of the equilibrium in a chemical reaction can be predicted with the aid of Le Chatelet's Principle in response to changes in temperature, concentration, or pressure. This is crucial, especially for industrial applications where it's crucial to predict and maximize yields.

According to Le Châtelet's principle, if a dynamic equilibrium is upset by changing the conditions, the equilibrium position will move to compensate for the change and restore the equilibrium.

To know about Le Chatelet's Principle from the link

brainly.com/question/2943338

#SPJ4

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The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate
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Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

            k = Ae^-(Ea/RT)

where,   k = rate constant

              A = pre-exponential factor

          Ea  = activation energy

             R = gas constant

              T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

         

      T1= (244 + 273.15) K = 517.15 K

      T2= (324+ 273.15) K =597.15 K

    K1  = 6.7 ,     K2 = ?

         R = 8.314 J/mol K

     Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

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