Question

The reaction shown below has a negative enthalpy change and a positive entropy change.

2O3 (g) yields 3O2 (g)

Which of the following best describes this reaction?

The reaction will only be spontaneous at high temperatures.
The reaction will only be spontaneous at low temperatures.
The reaction will not be spontaneous at any temperature.
The reaction will be spontaneous at any temperature.

Answer 1

@thomaster could you help me determine this?

Answer 2

@aaronq

Answer 3

@miraclemilktea2 could you help me?

Answer 4

@Abhisar

Answer 5

@iPwnBunnies

Answer 6

Do you know how spontaneity is determined with these thermodynamic values?

Answer 7

no i dont

Answer 8

the Gibbs free energy change (\(\Delta G\)) is indicative of the energy potentials in a dynamic equilibrium. The expression relating the enthalpy and entropy is:

\(\Delta G=\Delta H-T\Delta S\)

\(\Delta G<0\) reaction is spontaneous
\(\Delta G>0\) reaction is non-spontaneous

What you want is to know when the reaction will be spontaneous.

Answer 9

oh alright. what should i plug into the equation in order to determine it?

Answer 10

yep

Answer 11

so what do i plug in?

Answer 12

positive and negative signs

Answer 13

oh ok

Answer 14

The question is asking at what temperatures will \(\Delta G\) be negative.

Answer 15

so would the reaction be spontaneous

Answer 16

at what temperatures?

Answer 17

how do i know that?

Answer 18

\(ΔG=ΔH−TΔS\)

\(ΔG=-X−T(+Y)=-X-TY\)

will \(\Delta G\) ever be positive?

Note that T can only take on positive values (absolute scale)

Answer 19

no it can't be positive

Answer 20

exactly. So \(ΔG<0\) at all temperatures.

Answer 21

which means that it's always spontaneous

Answer 22

that makes sense that you very much! :)

Answer 23

thank*

Answer 24

i'm glad it does! no problem !