THREE QUESTIONS ON EQUILIBRIUM CONSTANT(Keq).

DROP YOUR ANSWERS INTO THE COMMENT BOX BELOW.

QUESTION 1.

If the value of the equilibrium constant K for the reaction represented as
A + B –> C is x.
What would be the value of the equilibrium constant K for this reaction below:-
C –> B + A, given that all the substances are in gaseous state.

QUESTION 2.

Consider the reaction:-
x + 3y –> 3z. Given the value of the equilibrium constant of the reaction at a certain condition to be equal to 0.4, find the equilibrium constant for the reaction:- z –> y + 3x

NOTE:- All are gaseous

QUESTION 3.

A system where Q > Keq, the solution is said to be ___ ?

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Akpan Favour Patrick
Akpan Favour Patrick
February 14, 2021 2:53 pm

For Question 1.) K = 1/x
For Question 2.) K= 1/(0.4)^1/3 = 1.36
For Question 3.) For a solution where Q>Keq, the solution is said to be producing more reactants.In other words, the solution is said to support the backward reaction

Edmond
Edmond
February 14, 2021 10:09 pm

You did well

Madu isaac C.
Madu isaac C.
February 15, 2021 7:25 pm

1. X=1/x
2. X=1/(x) ^1/3 i.e X= 1/(0.4)^1/3= 1.35
3. Supersaturated solution

Grãcious Grace
Grãcious Grace
February 15, 2021 9:27 pm

1, the equilibrium constant will be the inverse of the equilibrium constant of the first equation.ie 1/x
2,
3,It shows that the solution is said to be in aqueous solution not gaseous

Ray
Ray
February 17, 2021 7:20 am

1) 1/x
2) if x is 1/3 the answer will be 1.4
3) supper saturated

Divine Uranta
Divine Uranta
February 18, 2021 4:58 am

1. 1/x
2.(1/0.4)^-3
3. It shift to the left

Onyeonoro Olivia
Onyeonoro Olivia
February 18, 2021 6:16 am

1.1/k
2.1.36
3.backward reaction

Ehiagwina Victory
Ehiagwina Victory
February 19, 2021 5:45 am

1. K2 = 1/x
2. K2 = (0.4)^3, K2= 0.064
3. Q>K
The equilibrium position will shift to the left

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