Math, asked by lMrLovel, 1 month ago

Q1. SOLVE PLS

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Answered by pratiksengupta1507
0

Answer:

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Step-by-step explanation:

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Answered by PRINCE100001
5

Step-by-step explanation:

Question :

  • For a given exothermic reaction, Kp and K'p are the equilibrium constants at temperatures T₁ and T₂ respectively. Assuming that the heat of reaction is constant in temperature range between T₁ and T₂, it is readily observed that

Solution :

Given that,

  • Equilibrium constant at temperature T₁ = Kp
  • Equilibrium constant at temperature T₂ = K'p

We have to find the relation between Kp and K'p.

In order to solve this question, first of all we have to write relation between equilibrium constant and temperature.

We know that equilibrium constant changes with the change in temperature of thermodynamics system.

❖ The equilibrium constant at two different temperatures is given by

\dag\:\underline{\boxed{\bf{\gray{log\dfrac{K_2}{K_1}=\dfrac{\Delta H}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)}}}}†

K₁ denotes equilibrium constant at T₁ temperature and K₂ denotes equilibrium constant at T₂ temperature.

In this question, K₁ = Kp and K₂ = K'p

For exothermic reaction : ∆H = -ve

For endothermic reaction : ∆H = +ve

♦ Final temperature (T₂) will increase as the reaction is exothermic.

Therefore, T₂ > T₁

Hence,

\sf\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)

will be positive.

\sf:\implies\:log\dfrac{K'_p}{K_p}=\dfrac{-H}{2.303R}\cdot (+ve)

Finally overall term will be negative.

\sf:\implies\:log\dfrac{K'_p}{K_p}=(-ve)

So, we can say that

:\implies\:\underline{\boxed{\bf{\orange{K_p > K'_p}}}}

∴ Option (1) is the correct answer!

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