Author Question: At 2600 K, G = 775 kJ for the vaporization of boron carbide:B4C(s) 4 B(g) + C(s)Find G and ... (Read 295 times)

vicotolentino · **on:** Feb 15, 2020

Question 1

At high temperatures boron carbide vaporizes according to the equation

B4C(s) ⇌ 4 B(g) + C(s)

Which equation describes the relationship between ΔG° and ΔG for this reaction?
◦ ΔG = ΔG° + R T ln (pB ∙ [C]/[B4C])
◦ ΔG = ΔG° - 4 R T ln pB
◦ ΔG = ΔG° + 4 R T ln pB
◦ ΔG = ΔG° + R T ln pB

Question 2

At 2600 K, ΔG° = 775 kJ for the vaporization of boron carbide:

B4C(s) ⇌ 4 B(g) + C(s)

Find ΔG and determine if the process is spontaneous if the reaction vessel contains 4.00 mol B4C(s), 0.400 mol of C(s), and B(g) at a partial pressure of 1.0 × 10-5 atm. At this temperature, R T = 21.6 kJ.
◦ ΔG = -220 kJ; spontaneous.
◦ ΔG = -220 kJ; nonspontaneous.
◦ ΔG = -270 kJ; spontaneous.
◦ ΔG = -270 kJ; nonspontaneous.

ambernicolefink · **Reply #1 on:** Feb 15, 2020

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