[Mitch]

So the reaction for the formation of benzene is

6C(s,graphite) + 3H₂(g) --> C₆H₆(l) dH_f = 49.0 kJ/mol

Since the change in entropy for this process is negative and the heat of formation is positive, delta G for the reaction will always be positive. So, why does benzene exist? I know there is a huge flaw in my reasoning somewhere, but I can't figure it out at the moment.

[Borek]

1. Are you sure change in etropy is positive? Intiution tells me it is not, but then I am barely awaken ATM.

2. Even if so, dG = dH - TdS - see that minus sign?

Don't confuse thermodynamic stability with chemical reactivity.

3. It works both ways.

Standard disclaimer: I can be completely off

[Mitch]

Yes, sorry entropy is negative. So there is no way the above reaction can go? So, why does Benzene exist. Sorry going to bed can't write.

[Donaldson Tan]

Most Common Suggestions I Make on the Forums.
1. Always start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
3. Chemical properties can be found in the MSDS blocks to your left.

if dG for a process is negative, then the process is spontaneous.

Yes, sorry entropy is negative. So there is no way the above reaction can go? So, why does Benzene exist. Sorry going to bed can't write.

Benzene exists, but it definitely does not form through the reaction route suggested by the chemical equation:
6C(s,graphite) + 3H₂(g) --> C₆H₆(l) dH_f = 49.0 kJ/mol

[Mitch]

Okay just making sure.