Calculate the equilibrium constant at 25 oC for the reaction above?

For the reaction:
3C(s) + 4H2(g) = C3H8(g),
ΔSo = -269 J/K and ΔHo = -103.8 kJ.
Calculate the equilibrium constant at 25 oC for the reaction above.

A) 1.0
B) 3.7 x 1019
C) 1.4 x 104
D) 2.1 x 1031
E) 1.0 x 1017

1 Answer
May 17, 2018

The correct answer is C) #1.4 × 10^4#

Explanation:

We know that

#bb((1))color(white)(m)ΔG^@ = ΔH^@ "-" TΔS^@#

and

#bb((2))color(white)(m)ΔG^@ = "-"RTlnK#

We can combine these equations to get

#bb((3))color(white)(m) "-"RTlnK = ΔH^@color(white)(l) "-"color(white)(l) TΔS^@#

Thus

#color(blue)(bar(ul(|color(white)(a/a)lnK = "-"(ΔH^@)/(RT) + (ΔS^@)/Rcolor(white)(a/a)|)))" "#

In this problem,

#ΔH^@ = "-103.8 kJ·mol"^"-1"" #
#R color(white)(mm)= 8.314 "J·K"^"-1""mol"^"-1"#
#T color(white)(mm)= "25 °C = 298.15 K"#
#ΔS^@ color(white)(l)= "-269 J·K"^"-1""mol"^"-1"#

Thus,
#lnK = "-"("-103 800" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K")))) + ("-269" color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))))/(8.314 × color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1")))) = "41.7 - 32.36 = 9.52"#

#K = e^9.52 = 1.36 × 10^4#