Question

Explain where I did it wrong: The entropy change for the decomposition of 1 mol of KClO3 is 494J at 25celsius. Calculate the entropy value for O2 if the entropy value for KClO3 and KCl are 143 J/mol-K and 82.6 J/mol-K, repspectively? 2KClO3 --> 2KCl+3O2

494(2) = 2(82.6)+3x - 2(143)
Should I time 2 since 2 mol of KClO3?

Answer 1

`S^@ = "179 J·K"^"-1""mol"^"-1"`

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

`color(blue)(bar(ul(|color(white)(a/a)Δ_text(r)S^@ = sum_n nS_text(products)^@ - sum_m mS_text(reactants)^@color(white)(a/a)|)))" "`

Here's what you did wrong: You must multiply the `S^@` value of each reactant and product by a number of moles equal to its coefficient in the balanced equation.

Write the equation using 1 mol of `"KClO"_3`

#color(white)(mmmmmmmm)"KClO"_3"(s)" → "KCl(s)" + "³/₂O"_2"(g)"; color(white)(m)ul(Δ_text(r)S^@)#
`S^@"/J·K"^"-1""mol"^"-1": color(white)(ml)143 color(white)(mmmml)82.6color(white)(mmmll)x color(white)(mmmml) 494`

`(1×82.6 + 3/2x -1×143) color(red)(cancel(color(black)("J·K"^"-1"))) = 494 color(red)(cancel(color(black)("J·K"^"-1")))`

`82.6 + 3/2x -143 = 494`

`3/2x = 494 - 82.6 - 143 = 268.4`

`x = 2/3 × 268.4 = 179`

∴ `S^@ = "179 J·K"^"-1""mol"^"-1"`