Question

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

Answer 1

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`