Solving Spontaneous Reactions Homework: ΔG -270 kJ

Click For Summary
SUMMARY

The discussion focuses on calculating the Gibbs free energy change (ΔG) for the vaporization of boron carbide (B₄C) at 2600 K, where ΔG° is given as 775 kJ. The user attempted to compute ΔG using the equation ΔG = -RT*ln(Q) but encountered discrepancies in their calculations, arriving at ΔG = 1044 kJ instead of the expected -270 kJ. The error stemmed from misunderstanding the role of solids in the reaction quotient, as the activity of solids is considered to be 1 and thus excluded from Q.

PREREQUISITES
  • Understanding of Gibbs free energy and its significance in thermodynamics.
  • Familiarity with the reaction quotient (Q) and its calculation.
  • Knowledge of the ideal gas law and partial pressures.
  • Basic principles of chemical equilibrium and the role of solids in reactions.
NEXT STEPS
  • Study the concept of reaction quotients and their application in thermodynamic calculations.
  • Learn about the significance of activity coefficients in chemical reactions.
  • Explore the implications of temperature on Gibbs free energy and spontaneity.
  • Review the ideal gas law and its relationship with partial pressures in reaction systems.
USEFUL FOR

Chemistry students, thermodynamics enthusiasts, and anyone studying chemical equilibrium and Gibbs free energy calculations.

τheory
Messages
43
Reaction score
0

Homework Statement


At 2600 K, ΔG° = 775 kJ for the vaporization of boron carbide:
B_{4}C (s) ⇔ 4 B (g) + C (s)

Find ΔG and determine if the process is spontaneous if the reaction vessel contains 4.00 mol of B_{4}C (s), 0.400 mol of C (s), and B (g) at a partial pressure of 1.0 x 10^-5 atm. At this temperature, R T = 21.6 kJ

2. The attempt at a solution

I first tried to find the reaction quotient but quickly realized that they did not give me the volume for this vessel. Thus, I just plugged everything into the reaction quotient (products/reactants) and got Q = 1*10^-21.

Then I plugged that into the ΔG = -RT*ln(Q) equation:

ΔG = -RT*ln(Q)
ΔG = -(21.6 J * 1000)*ln(1*10^-21)
ΔG = 1.044*10^6 J = 1044 kJ

The answer on the back of the book says that ΔG is in fact -270 kJ; how is this the answer? What mistake did I do in my calculations?
 
Physics news on Phys.org
Please write the reaction quotient.
 
Q=\frac{(pB)^4*[C]}{[B_{4}C]}
 
What do you know about solids in this context?
 
Um all I can figure out is that the solids are within a closed vessel and that the temperature also appears to be 2600 K since 21.6kJ/8.314 = 2600. The only issue regarding the solid is that I don't know what the volume is, which is prohibiting me from finding the concentration mol/L.

The only other thing is that maybe I have the wrong idea of how concentration of a solid works?
 
Activity of solids is assumed to be 1, so they are not present in the reaction quotient.
 

Similar threads

Replies
1
Views
3K
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 131 ·
5
Replies
131
Views
10K