Respuesta :
Answer:
Q < K because K increased.
Reaction equilibrium shift towards right.
Explanation:
Reaction equilibrium: [tex]A\rightleftharpoons B[/tex]
For this reaction, [tex]\Delta H[/tex] is positive. This signifies that the given reaction is endothermic. So, with increase in temperature, reaction equilibrium shift towards right in accordance with Le-chatelier principle.
So, with increase in temperature, more A will be converted to B.
Equilibrium constant in terms of concentration, [tex]K_{c}=\frac{[B]}{[A]}[/tex] where [B} and [A} are equilibrium concentrations of B and A respectively.
Hence [tex]K_{c}[/tex] will increase with increase in temperature.
At equilibrium, Q (reaction quotient) = [tex]K_{c}[/tex] .
But immediately after an increase in temperature, Q < [tex]K_{c}[/tex] as [tex]K_{c}[/tex] is higher in this increased temperature as compared to previous temperature.
Equilibrium is defined as the state in which reactants and products are present in such a condition, where no further reaction can proceed. The reaction's equilibrium can be shifted to regulate the chemical reaction.
The correct answer is:
Option C. Q < K because K increased
As we know, reaction equlibrium is expressed as:
A [tex]\rightleftharpoons[/tex] B
The reaction is represented by [tex]\Delta[/tex]H as positive. The reaction can be represented as endothermic. Thus, when the temperature rises the reaction equilibrium shift towards the right.
The pattern follows the Equilibrium law.
The equilibrium constant in terms of concentration can be written as:
[tex]\text K_c&=[\dfrac{\text B}{\text A}][/tex]
where,
[B] = equilibrium concentration of B.
[A] = equilibrium concentration of A.
Thus, K[tex]_c[/tex] will increase in temperature, as Q = K[tex]_c[/tex].
Now, the rise in the temperature, the Q < K[tex]_c[/tex], as K[tex]_c[/tex] will be higher.
Therefore, option C is correct.
To know more about equilibrium law, refer to the following link:
https://brainly.com/question/16785869
