Question #baa36

1 Answer
Aug 7, 2016

Answer:

Here's what I get.

Explanation:

Step 1. Calculate the value of #ΔG# at 1600 K

#ΔG = "37 850 J" +"11.69 J"·color(red)(cancel(color(black)("K"^"-1"))) × 1600 color(red)(cancel(color(black)("K"))) = "(37 850 + 18 704) J" = "56 554 J"#

Step 2. Calculate the value of the equilibrium constant

#ΔG = "-"RTlnK#

#lnK = ("-"ΔG)/(RT) = ("-55 564" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 1600 color(red)(cancel(color(black)("K")))) = "-4.177"#

#K = e^"-4.177" = "0.015 34"#

Step 3. Calculate the pressure of #"CO"_2# at equilibrium

#"Ni" + "CO"_2 ⇌ "NiO" + "CO"#

#K_P = P_"CO"/P_"CO₂"#

#P_"tot" = P_"CO₂" + P_"CO" = "1 atm"#

Let #P_"CO" = xcolor(white)(l) "atm"#. Then #P_"CO₂" = (1-x) color(white)(l)"atm"#.

Then

#K_P = x/(1-x) = "0.015 34"#

#x = "0.015 34"(1-x) = "0.015 34 - 0.015 34"x#

#"1.015 34"x = "0.015 34"#

#x = "0.015 34"/"1.015 34" = "0.015 11"#

#1-x = "1 - 0.015 11" = 0.985#

∴ At equilibrium, #P_"CO₂" = "0.985 atm"#.

This is the minimum pressure of #"CO"_2# that will cause #"Ni"# to be oxidized.