Question

Calculate the temperature above which the reduction of lead oxide to lead in the following reaction become spontaneous
PbO(s) + C(s) Pb(s) + Co(g)
Given AH = 108.4 kJ mol . delta s = 190 JK-1 mol​

Answer 1

Answer:

We have,

ΔS = ΔH/T

So, T = ΔH/ΔS  

Here, ΔH = 108.4 kJ /mol= 108.4 x 103 J/mol

ΔS=  190 J/K mol

So, S = 108.4 x 103 / 190 = 570.5K  

Now, we see, if T is greater than 570.5K,  

TΔS is greater than ΔH and so ΔG = ( ΔH-TΔS) will be negative and the reaction becomes spontaneous.

Answer 2

Answer:

Explanation:

The formula of Entropy and Gibbs free reaction is:  

ΔS = ΔH/T But Since we want to find the temperature of lead oxide reduction of lead oxide leading to a spontaneous reaction it will be used as such:  

T = ΔH/ΔS

In the reaction, ΔH = 108.4 kJ /mol= 108.4 x 103 J/mol

ΔS=  190 J/K mol  

By putting values in their places we get:

S = 108.4 x 103 / 190 = 570.5K  

Hence if Temperature (T) is greater than 570.5K it becomes a spontaneous reaction. As TΔS is greater than ΔH and so ΔG = ( ΔH-TΔS) will be negative and the reaction becomes spontaneous