Question

Consider the reaction: A <=> B. Under standard conditions at equiliubrium, the concentrations of the compounds are [A] = 1.5 M, and [B] = 0.5 M. Keq' for this reaction is ____ and ∆G°' is _____. (You should not need a calculator for this.)

Answer 1

See explanation below

Step-by-step explanation:

In this case, let's write the equation again:

A <------> B Keq = ?

As we are using standard conditions, we can assume we have a temperature of 0 °C (273 K) and 1 atm.

To get the equilibrium constant we only do the following:

Keq = [B] / [A]

However, the problem is asking the reverse equilibrium constant (because of the ' in Keq'), so, we have to do the reverse division:

Keq' = [A]/[B]

Replacing the given values of A and B:

Keq' = 1.5/0.5 = 3

We have the equilibrium constant, we can calculate now the gibbs free energy with the following expression:

ΔG°' = -RTlnKeq'

As Keq' is > 1, the negative logaritm will result into a negative result or a number < 0, so, calculating this we have:

ΔG°' = -8.31 * 273 ln3

ΔG°' = -2.492.34 J

Answer 2

`Keq'>1\\\Delta G'<0`

Step-by-step explanation:

Hello,

In this case, for the given reaction, the equilibrium constant turns out:

`Keq=([B])/([A])=(0.5M)/(1.5M) =1/3`

Nonetheless, we are asked for the reverse equilibrium constant that is:

`Keq'=(1)/(Keq)=3`

Which is greater than one.

In such a way, the Gibbs free energy turns out:

`\Delta G'=-RTln(Keq')\\`

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.