Question

How can one estimate the value of K at temperatures other than 25°C for a reaction? How can one estimate the temperature where K = 1 for a reaction? Do all reactions have a specific temperature where K = 1?

Short answer

1. The temperature at which value of change in Gibbs free is equal to zero and K=1.
2. The Gibbs free value is greater than zero and K is less than oneif the change in enthalpy is positive and the change in entropy is negative.
3. The Gibbs free value is less than zero and K is greater than one if the change in enthalpy is negativeand the change in entropy is positive.

Step-by-step solution

Step 1: Introduction to the Concept

Entropyis a characteristic of a system that quantifies the unpredictability or degree of chaos. It's a state function, after all. With an increase in the universe's entropy, spontaneous change occurs.

The mathematical equation for the standard entropy value at room temperature is given as follows,

$\Delta {G^o}_{298}=\Delta H^o-T\Delta S^o$(or)

$\Delta G^o=\Delta {G^o}_{298}=\sum n{G^o}_{298}\left(Products\right)-\sum p{G^o}_{298}\left(Reactants\right)$

In the balanced chemical equation, n and p indicate the coefficients of reactants and products.

Step 2: Solution Explanation

With the help of the equilibrium constant, the term determines the equilibrium position(K).

At temperatures other than 25°C, K is always calculated by the following expression:

$\Delta G^o=-RTlnK$

To use the above expression,firstcompute the change in Gibbs free energy by assuming that enthalpy and entropy fluctuate independently of temperature.

The reaction is spontaneous in reverse reaction if the value of is

larger than zero and the equilibrium constant (K) is less than 1

, and it is spontaneous in forward reaction if the value of is less than zero and the equilibrium constant (K) is greater than 1.

The reaction is at equilibrium if $\Delta G^o$is equal to zero and the equilibrium constant (K) is equal to one.

$\Delta G^o=\Delta H^o-T\Delta S^o$

is the formulation for the relationship betweenchange in Gibbs free energy, change in enthalpy, change in entropy, and temperature.

For computing temperature, the put value of $\Delta G^o$is equal to zero:

$0=\Delta H°-T\Delta S°$

$T=\frac{\Delta H°}{\Delta S°}$

Step 3: Explanation of reasons

The above temperature expression is valid only when theenthalpy and entropy changes have the same sign.

If the signs of change in enthalpy and change in entropy vary, there is no temperature at which the value of change in Gibbs free is zero and K = 1.

If thechange in enthalpy is positive and the change in entropy is negative, theGibbs free value is more than zero and K is less than one.

If thechange in enthalpy is negative and the change in entropy is positive, theGibbs free value is less than zero and K is more than one.