If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Course: Chemistry library>Unit 13

Lesson 3: Gibbs free energy

Changes in free energy and the reaction quotient

The relationship between Gibbs free energy, standard Gibbs free energy, and the reaction quotient Q. Predicting the direction of the reaction when we don't have standard conditions.

Want to join the conversation?

• I have been keen to watch all the videos in the suggested order (Chemistry playlist). However, to understand this particular video, it seems that there is a missing one that explains why deltaG = deltaG0 + RTln(Q) ! Any suggestion where can I learn about this relationship between Gibbs energy and the reaction quotient?
• At 9.35 how did he know that K = 6.1 * 10^5?
How was it calculated?
• The equilibrium constant has been determined experimentally. For this video, the author looked up the number.

This reaction is called the Haber Process.
• The math on the last calculation seems off. When I calculated RTlnK I got 33004.2674. Why did he round this number to (33.0 x 10^3)? When he does that and adds (-33 x 10^3) the equation balances to zero, but why would he round mid-equation? If you don't round that number mid-equation you end up with deltaG = 4.27.
(1 vote)
• He didn't have to round off, but he would have gotten the same result.
Your precision is limited to 3 significant figures by the 298 K term (it's really 298.15 K).
So the best value you can get for RTlnQ is 33 000 J = 33.0 × 10³ J.
Δ G = -33.0×10³ J + 33.0×10³ J = 0.0 × 10³ J.
A value of 4.27 J is the same as 0.0 × 10³ J within the limits of uncertainty.
• For this equation for Gibbs Free Energy, dG= dG0 + RTlnQ, or dG=-RTlnK, will it matter what equilibrium constant, Kp or Kc, is used for gases? My reasoning is that Kp=Kc(RT)^n, so whichever K value is used, different Free Energy values would be calculated, right?
• Lets say we have an ideal gas A , then (Pa)V= nRT <-> n/V = (Pa)/RT = [A] , and the same for another gas B and C .

Supose we have this reaction x A + y B ---> z C

Kc = [A]^x . [B]^y / [C]^z = (Pa/RT)^x . (Pb/RT)^y / (Pc/RT)^z

Kc = (Pa^x) . (Pb^y) / (Pc)^z . (1/RT)^x . (1/^RT)^y . (1/RT)^-z

Kc = Kp . (1/RT)^(x+y-z)

I think in this way.
• How do you figure that kp out?
• How did you know the value of kp in the second question (written in red marker)
• Why is the gas constant used even when the concentrations are in molarity?
• It doesn't matter what inputs we use for the equilibrium constant is in since it is unitless. The values whose units we care about when using the gas constant are the free energy and temperature.

Hope that helps.
• At about , why do we directly plug in the Q of partial pressures? Isn't the Q in the equation supposed to be the molar Q, not the partial pressure one?

So shouldn't we convert Q_p to Q of moles using PV = nRT?
• The equation at can take either Qc (using molarity) or Qp (using partial pressures). No need to convert.

Hope that helps.