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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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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

You did well

1. X=1/x

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

3. Supersaturated solution

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

1) 1/x

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

3) supper saturated

1. 1/x

2.(1/0.4)^-3

3. It shift to the left

1.1/k

2.1.36

3.backward reaction

1. K2 = 1/x

2. K2 = (0.4)^3, K2= 0.064

3. Q>K

The equilibrium position will shift to the left