gen chem questions

[kiron]

1. can someone explain to me the convention system for pos. and neg. w in

E(internal) = q + w for adiabatic and isothermal processes?

[Avinash]

The key thing to remember is that work done by the system (expansion) reduces internal energy and, work done on the system (compression) increases internal energy.

 

With that in mind, if we define work as W=P(delta)V, then when expansion occurs (delta V is positive), then work is positive. When compression occurs (delta V is negative), then work is negative.

 

(delta)E = Q - W

 

As you can see, when work is done by the system (expansion), W is positive, and internal energy is decreased. When work is done on the system (compression), W is negative, and internal energy is increased.

 

In adiabatic systems, there is no heat exchange, meaning Q=0.

(delta)E = 0 -W = -W

 

Isothermal systems, there is no change in temperature of the system, meaning no change in internal energy, since we say internal energy is a direct reflection of the temperature.

 

(delta)E = 0 = Q - W

0 = Q - W

W = Q

 

According to this, any energy given off by the system in the form of heat, must equal energy coming into the system in the form of work (compression). Any energy coming into the system in the form of heat, must equal energy leaving the system in the form of work (expansion).

[tallguy81]

Another Chem Question,

 

When Keq is Positive, is it safe to definitely say that Free Energy (G) is negative, and therefore has to be spontaneous, via the equation G=-RTlnK? Where does the reaction quotient (Q) come into play? I thought if:

 

Q < K : Products favored, therefore spontaneous

Q > K : Reactants favored, therefore non-spontaneous

 

Or does spontaneity have nothing to do with position of equilibrium?

[kiron]

So E(internal) = Q - W?

I always thought E = Q+W or is it all relative?

[MoSquared]

Another Chem Question,

 

When Keq is Positive, is it safe to definitely say that Free Energy (G) is negative, and therefore has to be spontaneous, via the equation G=-RTlnK? Where does the reaction quotient (Q) come into play? I thought if:

 

Q < K : Products favored, therefore spontaneous

Q > K : Reactants favored, therefore non-spontaneous

 

Or does spontaneity have nothing to do with position of equilibrium?

 

Keq is always positive isn't it. Can you get a negative Keq??? Keq=[Products]^x/[Reactants]^y can you have a negative concentration???

 

Or do you mean if lnK is negative?

[Avinash]

So E(internal) = Q - W?

I always thought E = Q+W or is it all relative?

 

E= Q + W if you define work as W=-P(delta)V

[gamma]

Keq is always positive isn't it. Can you get a negative Keq??? Keq=[Products]^x/[Reactants]^y can you have a negative concentration???

 

Or do you mean if lnK is negative?

 

you're right; K or Q can never be negative. tallguy81 probably meant ln K

 

I believe you can generalize to say that a large K => - G, small K => + G

If you go back to G = -RTlnK:

when K is larger than 1, ln K is larger than 1 and positive, and you get -G

when K is smaller than 1, ln K is less than 1 and negative, and you get +G

keep in mind that when K =1 you're at equilibrium ( delta G = delta Gnaught +RT ln Q)

 

though K and G are both thermodynamic properties, I don't really know if you could say whether the equilibrium determines spontaneity...

[Avinash]

Standard state is defined as the conditions where the concentration of all reactants and products are 1M. With that in mind, the reaction quotient Q, is always 1 at the standard state, for any reaction.

 

dG0 is the change in gibbs energy as we go from the standard state (all concentrations=1 and Q=1) to the equilibrium state (Q=Keq).

 

dG0 = -RTlnKeq

 

According to this equation, any reaction with a Keq greater than 1 is spontaneous as it moves from standard state (1M concentration of products and reagents) to establish the equilibrium state. This is because the ln of any number greater than 1 is positive, meaning dG0 will be negative.

 

However, for real world purposes, we would like to gauge the spontaneity of reactions starting from a position Q different from standard state. For this, we would look at the dG of the reaction.

 

dG = dG0 + RTlnQ

in other words

dG = -RTlnKeq + RTlnQ

 

According to this, if Q > Keq, then RTlnQ will be large enough to make dG positive, or non spontaneous.

If Q < Keq, then dG will be negative, and the reaction will be spontaneous.