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

Answer:

$Keq'>1\\\Delta G'$

Explanation:

Hello,

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

$Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3$

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

$Keq'=\frac{1}{Keq}=3$

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.

Answer 2

Answer:

See explanation below

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