Answer:
- <u><em>Option a. 6200 K</em></u>
Explanation:
<u>1) Data:</u>
- V₁ = 0.66 liter
- P₁ = 42.9 mmHg
- T = 261.2 K
- T₂ = ?
- V₂ = 7.63 liter
- P₂ = 872.15 mmHg
<u>2) Formula:</u>
Combined law of gases:
<u>3) Solution:</u>
T₂ = P₂ V₂ T₁ / (P₁ V₁)
T₂ = 872.15 mmHg × 7.63 liter × 261.2 K / ( 424.9 mmHg × 0.66 liter)
T₂ = 6198 K
- Rounding to 2 significant figures, that is 6200 K, which is the first choice.
Answer:

Explanation:
In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):

Being Keq:
![K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5Bfructose%5D%5BPi%5D%7D%7B%5BFructose-1-P%5D%7D)
Initial conditions:
![[Fructose-1-P]=0.2M](https://tex.z-dn.net/?f=%5BFructose-1-P%5D%3D0.2M)
![[Fructose]=0M](https://tex.z-dn.net/?f=%5BFructose%5D%3D0M)
![[Pi]=0M](https://tex.z-dn.net/?f=%5BPi%5D%3D0M)
Equilibrium conditions:
![[Fructose-1-P]=6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BFructose-1-P%5D%3D6.52%2A10%5E%7B-5%7DM)
![[Fructose]=0.2M-6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BFructose%5D%3D0.2M-6.52%2A10%5E%7B-5%7DM)
![[Pi]=0.2M-6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BPi%5D%3D0.2M-6.52%2A10%5E%7B-5%7DM)


Free-energy for T=298K (standard):

