Question

For the process 2SO2(g) + O2(g) --> 2SO3(g),

ΔS = –187.9 J/K and ΔH = –198.4 kJ at 297.0 K are known.

What is the entropy of this reaction? Use ΔG = ΔH – TΔS

Answer 1

c

Step-by-step explanation:

gt

Answer 2

–187.9 J/K

Step-by-step explanation:

The equation that relates the three quantities is:

`\Delta G = \Delta H - T \Delta S`

where

`\Delta G` is the Gibbs free energy

`\Delta H` is the change in enthalpy of the reaction

T is the absolute temperature

`\Delta S` is the change in entropy

In this reaction we have:

ΔS = –187.9 J/K

ΔH = –198.4 kJ = -198,400 J

T = 297.0 K

So the Gibbs free energy is

`\Delta G=-198,400-(297.0)(-187.9)=254.2 kJ`

However, here we are asked to say what is the entropy of the reaction, which is therefore

ΔS = –187.9 J/K